hypothesis for empirical formula

Empirical Formula: Definition and Examples

How to read the element ratio in an empirical formula

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• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

The empirical formula of a compound is defined as the formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols.

Also Known As: The empirical formula is also known as the  simplest formula  because the subscripts are the smallest whole numbers that indicate the ratio of elements.

Key Takeaways: Empirical Formula

• The empirical formula is also known as the simplest formula in chemistry.
• It gives the smallest whole number ratio of elements in a compound using subscripts following element symbols.
• In some cases, the empirical formula is the same as the molecular formula, which gives the actual number of atoms in a compound (e.g., H 2 O).
• Otherwise, the molecular formula is a multiple of the empirical formula (e.g., CH 2 O is the empirical formula for glucose, C 6 H 12 O 6 ).

Empirical Formula Examples

Glucose has a molecular formula of C 6 H 12 O 6 . It contains 2 moles of hydrogen for every mole of carbon and oxygen. The empirical formula for glucose is CH 2 O.

The molecular formula of ribose is C 5 H 10 O 5 , which can be reduced to the empirical formula CH 2 O.

How to Determine Empirical Formula

• Begin with the number of grams of each element, which you usually find in an experiment or have given in a problem.
• To make the calculation easier, assume the total mass of a sample is 100 grams, so you can work with simple percentages. In other words, set the mass of each element equal to the percent. The total should be 100 percent.
• Use the molar mass you get by adding up the atomic weight of the elements from the periodic table to convert the mass of each element into moles.
• Divide each mole value by the small number of moles you obtained from your calculation.
• Round each number you get to the nearest whole number. The whole numbers are the mole ratio of elements in the compound, which are the subscript numbers that follow the element symbol in the chemical formula.

Sometimes determining the whole number ratio is tricky and you'll need to use trial and error to get the correct value. For values close to x.5, you'll multiply each value by the same factor to obtain the smallest whole number multiple. For example, if you get 1.5 for a solution, multiply each number in the problem by 2 to make the 1.5 into 3. If you get a value of 1.25, multiply each value by 4 to turn the 1.25 into 5.

Using Empirical Formula to Find Molecular Formula

You can use the empirical formula to find the molecular formula if you know the molar mass of the compound. To do this, calculate the empirical formula mass and then divide the compound molar mass by the empirical formula mass. This gives you the ratio between the molecular and empirical formulas. Multiply all of the subscripts in the empirical formula by this ratio to get the subscripts for the molecular formula.

Empirical Formula Example Calculation

A compound is analyzed and calculated to consist of 13.5 g Ca, 10.8 g O, and 0.675 g H. Find the empirical formula of the compound.

Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H.

13.5 g Ca x (1 mol Ca / 40.1 g Ca) = 0.337 mol Ca

10.8 g O x (1 mol O / 16.0 g O) = 0.675 mol O

0.675 g H x (1 mol H / 1.01 g H) = 0.668 mol H

Next, divide each mole amount by the smallest number or moles (which is 0.337 for calcium) and round to the nearest whole number:

0.337 mol Ca / 0.337 = 1.00 mol Ca

0.675 mol O / 0.337 = 2.00 mol O

0.668 mol H / 0.337 = 1.98 mol H which rounds up to 2.00

Now you have the subscripts for the atoms in the empirical formula:

Finally, apply the rules of writing formulas to present the formula correctly. The cation of the compound is written first, followed by the anion. The empirical formula is properly written as Ca(OH) 2

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the formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula

the formula of a compound in which the subscripts give the actual number of each element in the formula

This formula is already at the smallest whole-number ratio. You can't reduce the H 2 O subscripts to anything smaller that is still a whole-number. What about HO 0.5 ? After all, 1 to 0.5 is still a 2:1 ratio. Ah, but it's not a whole number ratio. That means H 2 O is the empirical formula for water. (It will also be the molecular formula.) CO 2 , NH 3 , CH 4 . There are the empirical formulas for carbon dioxide, ammonia, and methane. They are also the molecular formulas.

This formula has two carbons, four hydrogens and two oxygens. So we could write the formula like this: C 2 H 4 O 2 See how there is a common factor of 2 among the subscripts? That means you are NOT looking at the empirical formula. You're looking at the molecular formula. Take out the factor to get the empirical formula of CH 2 O.

Substance Molecular Formula Empirical Formula Scaling Factor Lactic acid CH 3 CH(OH)COOH CH 2 O 3 Erythrose C 4 H 8 O 4 CH 2 O 4 Ribose C 5 H 10 O 5 CH 2 O 5 Glucose C 6 H 12 O 6 CH 2 O 6 See how you take out the scaling factor to get the empirical? C 4 H 8 O 4 has a common factor of 4 among the subscripts. Take it out to get the empirical formula of CH 2 O. Notice two things: 1. The molecular formula and the empirical formula can be identical. 2. You scale up from the empirical formula to the molecular formula by a whole number factor.

Yes, there is a compound with the formula CH 2 O. Its name is formaldehyde. The empirical formula is CH 2 O since there are no common factors (other than 1) in the subscripts. CH 2 O is the molecular formula because the subscripts give the actual number of atoms of each element in the molecule.

Percent to mass Mass to mole Divide by small Multiply 'til whole

Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen.

for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N

for Mg: 2.97 mol / l.99 mol = 1.49 for N: 1.99 mol / l.99 mol = 1.00

for Mg: 2 x 1.49 = 2.98 (i.e., 3) for N: 2 x 1.00 = 2.00

carbon: 68.54 grams hydrogen: 8.63 grams oxygen: 22.83 grams

carbon: 68.54 / 12.011 = 5.71 mol hydrogen: 8.63 / 1.008 = 8.56 mol oxygen: 22.83 / 16.00 = 1.43 mol

carbon: 5.71 ÷ 1.43 = 3.99 hydrogen: 8.56 ÷ 1.43 = 5.99 oxygen: 1.43 ÷ 1.43 = 1.00

C 4 H 6 O gives an "EFW" of 70.092.

140 ÷ 70 = 2

C 4 H 6 O times 2 gives a formula of C 8 H 12 O 2 .

Molecular Formula vs Empirical Formula

Core Concepts

In this tutorial, you will learn what an empirical formula and molecular formula are, and the differences of molecular formula vs empirical formula. You will also learn how to convert between the molecular and empirical formula. Lastly, you’ll learn how to use an empirical formula to find a molecular formula with data.

Topics covered in other articles

• Molecules / Compounds
• Chemical Bonding
• Structural Isomers
• Percent composition
• How to calculate percent yield

molecule : A compound formed through the bonding of 2 or more atoms

n-value: an integer that is multiplied by the empirical formula in order to obtain the molecular formula

Molecular Formula

A molecular formula represents the number of each atom present in a given molecule . Essentially, it is a list of what elements are found in a molecule, and how many there are of each.

The molecular formula can give useful information about the properties of a molecule. It is the most common way to describe simple molecules beside their names.

Molecular formulas don’t always show the full story – because they only list the identity and number of elements in a molecule, the structure can sometimes be ambiguous. Some compounds have the same molecular formula (meaning they have the same atoms in the same quantities) but are arranged differently. Compounds with the same formula but different shapes or connectivity between atoms are called isomers.

C 6 H 12 O 6 → The molecular formula used to describe fructose, glucose, and galactose.

N 2 O 4 → The molecular formula used to describe nitrogen tetroxide.

C 5 H 3 N 3 → The molecular formula used to describe cyanopyrazine.

Empirical Formula

The empirical formula represents the relative amount of the elements in a molecule. This is sometimes different than the molecular formula, which gives the exact amounts. Empirical formulas are useful because knowing the relative amount of every element in a molecule can be extremely helpful for determining the molecular formula.

Here is a simple explanation:

An empirical formula is a way of expressing the composition of a chemical compound. It shows the relative proportions of the different elements that make up the compound, but not the actual numbers or amounts of atoms. For example, the empirical formula for glucose (a type of sugar) is CH 2 O, which tells us that it is made up of carbon, hydrogen, and oxygen, but does not give us any information about the exact number of atoms of each element.

To find the empirical formula of a compound, you need to determine the relative proportions of each element in the compound, and then express those proportions as a simplified formula. This can be done by analyzing the compound’s chemical properties or by performing experiments.

In order to determine the true number of each atom in a molecule, it is important to obtain an n-value . The n-value is a whole number that the empirical formula is multiplied by in order to obtain the molecular formula. The molecular formula and empirical formula can sometimes be the same, as long as the ratio of atoms in the molecular formula is at its simplest. In other words, if the n-value is 1.

CH 2 O → The empirical formula of fructose, glucose, and galactose once reduced.

NO 2 → The empirical formula of nitrogen tetroxide (N 2 O 4 ) once reduced.

C 5 H 3 N 3 → The empirical formula and molecular formula for cyanopyrazine are the same, as the ratio of the atoms cannot be simplified.

What’s the difference?

Remember that the molecular formula is a list – it represents each and every atom found in a molecule. Molecular formulas are used to communicate about molecules in many settings, and can sometimes give insight on the compound’s characteristics.

The empirical formula does not necessarily tell us how many atoms there are of each element in a molecule. It only shows the proportions between them. The empirical formula is mainly used in experimental settings, where it acts as a stepping stone to the molecular formula.

In rare cases, the empirical formula can be useful on its own. For example, some organic chemists use the carbon-to-oxygen ratio from an empirical formula to quickly estimate how reactive a chemical is. Generally though, empirical formulas are not useful for understanding the properties of molecules. In some cases, empirical formulas will not even be possible to treat as molecular formulas. For example, benzene (C 6 H 6 ) has the empirical formula CH. CH is not a molecule that could actually exist – this goes to show that while the empirical formula is a useful tool to find some information, it should not be used to make conclusions about the behavior of compounds it represents.

Take the formula CH 2 O.

First, consider it as a molecular formula. There is only one way to build a molecule with that formula. That chemical is called formaldehyde – it’s very toxic but also very useful in reactions, as an embalming agent, and elsewhere.

Now consider CH 2 O as an empirical formula. What molecular formulas could it represent? Formaldehyde, acetic acid (vinegar), lactic acid (an important part of cellular respiration), and glucose all share the empirical formula CH 2 O.

These molecules are all extremely different, ranging from a simple sugar to a dangerous carcinogen. While they all have the same empirical formula, their molecular formulas and characteristics differ greatly.

How to convert a molecular formula to its empirical formula:

• Let’s start with a compound, for example ethyl acetate : C 4 H 8 O 2
• Find the greatest common factor (GCF) between the number of each atom. In this case, the GCF between 2, 4, and 8 is 2, meaning 2 is the n-value.
• Divide the number of each atom by the greatest common factor (AKA the n-value). In this case C= 4/2= 2 , H= 8/2= 4, O= 2/2= 1.
• Replace the previous values of each atom with the newly calculated ones. This would result in the empirical formula of C 2 H 4 O.

How to find a molecular formula using its empirical formula

While finding the empirical formula from the molecular formula can be a little tricky, doing the opposite is extremely easy. You simply multiply each element’s subscript in the empirical formula by the n-value. Here’s an example question:

“What molecular formula corresponds to the empirical formula C 3 H 4 N 2 and an n-value of 3?”

To solve it, we multiply each atom’s subscript by the n-value:

It might seem strange that the n-value is specified. The reason we need the n-value to find the answer is that there are, in theory, an infinite number of molecular formulas that share the empirical formula C 3 H 4 N 2 , one for every value of n. Therefore, we need to know “where we’re going” beforehand.

Finding molecular formulas from data

In general, the procedure for finding molecular formulas from experimental data will be as follows:

• Write the moles of each element in the sample. These can be found from experimental data like reaction products, percent composition, or otherwise.
• Find the element with the least moles. Divide the moles of every element by that amount. This will give the ratio of each element to the one with the least moles. These ratios may not be whole numbers, e.g. 1.33, 2.5, etc.
• If the relative amounts are all whole numbers, skip this step. Otherwise, pick one number to multiply every relative amount of each element by so that they become whole numbers. For example, C = 1.67, H = 2.33, O = 1 would be transformed to C = 5, H = 8, O = 3 by multiplying through by 3.
• Write a formula with subscripts equal to the numbers obtained in the last step. The example above would be written C 5 H 8 O 3 . You’re done! That’s the empirical formula.

For More Help, Watch Our Interactive Video on Molecular and Empirical Formulas!

Practice problem – an empirical formula experiment.

Below is an example of how one can find the molecular formula with experimental data by using the empirical formula.

A chemist receives a canister of an unknown chemical, and she needs to figure out its molecular formula. She knows already that the chemical only contains carbon and hydrogen.

The researcher takes a 50.0 gram sample of the unknown compound and burns it in a combustion reaction with excess oxygen gas, creating CO 2 and H 2 O. After collecting the products, she finds that the reaction produced 156.8 g CO 2 and 64.2 g H 2 O.

Finding the empirical formula

The chemist first finds the moles of each product:

Since all of the carbon in CO 2 came from the original molecule, as did all of the hydrogen in water, the chemist further finds:

There is one relative mole of carbon and two relative moles of hydrogen. This means that the empirical formula of the starting molecule is CH 2 .

Finding the molecular formula

After some more testing, the chemist concludes that the molecular weight of the unknown chemical is 42.09 grams per mole. She finds the empirical weight (that is, the weight of the empirical formula of the compound, CH 2 ) to be 14.03 grams per mole. By definition, the n-value times the empirical formula equals the molecular formula. She uses this relationship to find the following:

With an n-value of 3, the chemist “multiplies” the empirical formula by 3 to find the molecular formula:

The chemist therefore concludes that the molecular formula of the unknown compound is C 3 H 6 .

If you follow the steps in this tutorial, any empirical formula problem should be a breeze. Remember that many compounds will share the same empirical formula, and finding the differences based on mass, physical properties, reactivity, and other characteristics is key to discovering the molecular formula.

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General Chemistry (Alexander Antonopoulos)

Lab 1: determining the empirical formula of a compound.

• Lab 2: Determine the Percentage of Water in a Hydrate
• Lab 16: Gravimetric Determination of a Precipitate
• Lab 9: Mole Relationships
• Lab 14a: Separation and Analysis of Cations
• Lab 14b: Analysis of Anions
• Lab 18: Separation by Liquid Chromatography
• Lab 17a: Absorption Spectra
• Lab 17b: Colorimetric Analysis
• Lab 6a: Standardizing a Solution of Sodium Hydroxide
• Lab 7a: Acid-Base Titration
• Lab 33: Determination of Calcium Carbonate Content of and Anti-Acid Pill
• Lab 29: Exploring Gas Laws
• Lab 27: Identifying an Unknown Metal
• Lab 28: Molecular Interaction in Ethanol and Acetone
• Lab 26: Conductometric Titration
• Lab 25: Order of Reaction
• Introduction
• Method and Equipment
• Data Analysis and Conclusion

Lab 1: Determining the Empirical Formula of a Compound:

The goal of this experiment is to determine the Empirical Formula of a Compound. (The Empirical Formula of a Compound is the simplest whole number ratio between the elements of a compound) If one can synthesize a compound from elements, then it is possible to determine an experimental empirical formula for the compound, from its molar and stoichiometric ratios.

I hypothesize that by reacting Mg with pure O, (or air, and then using water and heat to remove any magnesium nitride formed upon exposure to the air), and then calculating the ratio between the moles of Mg and O, I should be able to determine experimentally the empirical formula of MgO.

Equipment and Materials :

• Crucible with lid
• Clay Triangle
• Bunsen Burner
• Glass Rod (for stirring)
• Mesh Pad, with Asbestos
• Bottle of Deionized Water

Procedure :

• Retrieved equipment, and assembled it properly at workstation. Inspected crucible and lid for defects. Measured mass of crucible with lid: 38.5987g
• Heated crucible (henceforth, assume that the lid is included, unless otherwise mentioned) with Bunsen Burner, heating it for 15 minutes, before removing crucible with the tongs, placing it upon the mesh pad, and allowing it to return to room temperature. All subsequent handling of the crucible was performed with the tongs. Measured mass of fired crucible: 38.5980g
• Measured mass of magnesium: 0.1867
• Measured mass of magnesium and crucible: 38.7860g
• Heated the crucible and magnesium on Bunsen Burner, until the crucible was red hot, then briefly removed the lid to introduce air to the magnesium. Heated crucible for 3 more minutes, before removing from heat, placing on mesh pad, and allowing it to cool to room temperature.
• Stirred the magnesium with the glass rod, before returning it to heat until red hot. After exposing the magnesium to air for the second time, removed it from the Bunsen Burner, placed it on the mesh pad, and allowed it to return to room temperature.
• Added 3 drops of deionized water to the crucible, before returning it to the Bunsen Burner. Heated the crucible for 15 minutes, opening the lid at points to allow the water vapor to escape. Removed crucible, placed it on the mesh pad, and allowed it to return to room temperature. Measured mass of crucible and Magnesium Oxide: 38.8873g
• Disposed of the MgO within the proper receptacle, then scrubbed the crucible, and reinspected it, before returning it to the drying station. Returned all equipment to its point of origin.

Data Analysis:

Determined mass of magnesium oxide by subtracting the mass of the crucible and lid, from the mass of the crucible, lid, and magnesium oxide: 0.2893g

Determined mass of oxygen by subtracting the mass of the magnesium from the magnesium oxide: 0.1013g

Determined amount of magnesium in the compound, by dividing the mass of the magnesium in the compound, by the number of grams per mole in 1 mole of magnesium: 0.0077mol

Determined amount of oxygen in the compound, by dividing the mass of the oxygen in the compound, by the number of grams per mole in 1 mole of oxygen: 0.0063mol

Used the molar ratio to determine the experimental empirical formula of the compound:

Taken to the nearest whole molecule, this experiment finds the empirical formula of MgO to be 1:1

Conclusion :

The goal of this experiment is to determine the Empirical Formula of a Compound. (The Empirical Formula of a Compound is the simplest whole number ratio between the elements of a compound) If one can synthesize a compound from elements, then it is possible to determine an experimental empirical formula for the compound, from its molar and stoichiometric ratios. By reacting Mg with pure O, (or air, and then using water and heat to remove any magnesium nitride formed upon exposure to the air), and then calculating the ratio between the moles of Mg and O, it is possible to determine experimentally the empirical formula of MgO.

The accepted empirical formula of MgO: 1:1

The experimental empirical formula of MgO: 1:1.221

Percent error: 22.1%

Possible sources of error include:

• Failure to properly react all of the Magnesium with Oxygen, and/or failure to react all of the Magnesium Nitride with water. (Disrupting subsequent measurements due to differences in mass)
• Failure to properly clean the paperclip used for mixing. (Thus removing from the contents of the crucible, and disrupting subsequent measurements)
• Human error is always in effect, given that the laboratory does not function under ideal conditions. As such, there is always the possibility of inaccuracies with measurement, perception of measurement, inaccuracies of equipment, and other such errors. (However, this is not likely to be the sole cause of the inaccuracies within this experiment, though it may contribute to it.)

Possible improvements that could be made for subsequent experiments include using more precise instruments, reacting the Magnesium in a pure Oxygen environment (removing the possibility of Magnesium Nitride forming, and possibly not reacting fully with water), and repeating the test multiple times, to minimize the possibility of an anomalous result skewing the final measurement.

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Finding the formula of hydrated copper(II) sulfate

In association with Nuffield Foundation

In this experiment, a known mass of hydrated copper(II) sulfate is heated to remove the water of crystallisation

This is a class experiment suitable for students who already have a reasonable understanding of the mole concept.

The mass of water is found by weighing before and after heating. This information is used to find x in the formula CuSO 4 .xH 2 O, using mole calculations

The degree to which the mole calculations need to be structured will depend on the ability and mathematical competence of the class. The outline structure given in the Procedure above is intended for students with reasonable mathematical competence and experience of mole calculations.

Given adequate access to top-pan balances, and skill in their use, students should be able to complete the experimental work in 30–40 minutes.

• Eye protection
• Crucible (note 1)
• Crucible tongs (note 2)
• Pipe clay triangle
• Bunsen burner
• Heat resistant mat
• Top-pan balance (± 0.01 g)

Apparatus notes

• Crucibles may be of porcelain, stainless steel or nickel, of capacity about 15 cm 3 , and should sit safely in the pipe clay triangles provided.
• Crucible tongs should have a bow in the jaws of the right size to pick up the hot crucibles safely.
• Hydrated copper(II) sulfate (HARMFUL, DANGEROUS FOR THE ENVIRONMENT), 2–3 g

Health, safety and technical notes

• Read our standard health and safety guidance
• Hydrated copper(II) sulfate, CuSO 4 .5H 2 O(s), (HARMFUL, DANGEROUS FOR ENVIRONMENT) – see CLEAPSS Hazcard  HC027c . The copper(II) sulfate should be provided as fine crystals. If large crystals are used, these should be ground down before use by students.
• Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulfate. Record all weighings accurate to the nearest 0.01 g.
• Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.
• Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.
• Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.
• Re-weigh the crucible and contents once cold.
• Calculation:
• Calculate the molar masses of H 2 O and CuSO 4  (Relative atomic masses: H=1, O=16, S=32, Cu=64)
• Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment
• Calculate the number of moles of anhydrous copper(II) sulfate formed
• Calculate the number of moles of water driven off
• Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed
• Write down the formula for hydrated copper(II) sulfate.

Source: Royal Society of Chemistry

Hydrated copper(II) sulfate apparatus set-up

Teaching notes

Remind students to zero (tare) the balance before each weighing.

Students will probably also have to be reminded about the need to allow the crucible and contents to cool thoroughly before weighing.

Metal crucibles (stainless steel or nickel) are much less vulnerable than porcelain crucibles.

Additional information

This is a resource from the  Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry. This collection of over 200 practical activities demonstrates a wide range of chemical concepts and processes. Each activity contains comprehensive information for teachers and technicians, including full technical notes and step-by-step procedures. Practical Chemistry activities accompany  Practical Physics  and  Practical Biology . 

© Nuffield Foundation and the Royal Society of Chemistry

• 11-14 years
• 14-16 years
• 16-18 years
• Practical experiments
• Quantitative chemistry and stoichiometry

Specification

• (j) concept of stoichiometry and its use in calculating reacting quantities, including in acid-base titrations
• (p) how to calculate the formula of a compound from reacting mass data
• 1.7.10 demonstrate knowledge and understanding that water of crystallisation can be removed by heating to constant mass and any thermal decomposition may be carried out to completion by heating to constant mass;
• 1.7.11 calculate the relative formula mass of compounds containing water of crystallisation;
• 1.7.12 calculate the percentage of water of crystallisation in a compound;
• 1.7.13 determine the empirical formulae of simple compounds and determine the moles of water of crystallisation present in a hydrated salt from percentage composition, mass composition or experimental data; and
• 2.6.2 demonstrate knowledge and understanding that water of crystallisation can be removed by heating to constant mass and any thermal decomposition may be carried out to completion by heating to constant mass;
• 2.6.3 calculate the relative formula mass of compounds containing water of crystallisation;
• 2.6.4 determine the empirical formulae of simple compounds and determine the moles of water of crystallisation present in a hydrated salt from percentage composition, mass composition or experimental data;
• Using mass of substance, Mᵣ, and amount in moles.
• d) the terms anhydrous, hydrated and water of crystallisation and calculation of the formula of a hydrated salt from given percentage composition, mass composition or based on experimental results
• The balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles and converting the numbers of moles to simple whole number ratios.
• Students should be able to balance an equation given the masses of reactants and products.
• WS.4.6 Use an appropriate number of significant figures in calculation.
• WS.2.4 Carry out experiments appropriately having due regard for the correct manipulation of apparatus, the accuracy of measurements and health and safety considerations.
• WS.2.6 Make and record observations and measurements using a range of apparatus and methods.
• WS.2.7 Evaluate methods and suggest possible improvements and further investigations.
• 1c Use ratios, fractions and percentages.
• 2a Use an appropriate number of significant figures.
• Deduce the stoichiometry of an equation from the masses of reactants and products and explain the effect of a limiting quantity of a reactant.
• Explain how the mass of a given substance is related to the amount of that substance in moles and vice versa.
• WS4.6 Use an appropriate number of significant figures in calculation.
• WS2.4 Carry out experiments appropriately having due regard for the correct manipulation of apparatus, the accuracy of measurements and health and safety considerations.
• WS2.6 Make and record observations and measurements using a range of apparatus and methods.
• WS2.7 Evaluate methods and suggest possible improvements and further investigations.
• 1c Use ratios, fractions and percentages
• 2a Use an appropriate number of significant figures
• 2d Carry out experiments appropriately having due regard for the correct manipulation of apparatus, the accuracy of measurements and health and safety considerations
• 2g Evaluate methods and suggest possible improvements and further investigations
• 4f Use an appropriate number of significant figures in calculation
• 1.51 Calculate the number of: moles of particles of a substance in a given mass ofthat substance and vice versa; particles of a substance in a given number of moles of that substance and vice versa; particles of a substance in a given mass of that…
• 1.53 Deduce the stoichiometry of a reaction from the masses of the reactants and products
• 1.51 Calculate the number of: moles of particles of a substance in a given mass ofthat substance and vice versa; particles of a substance in a given number of moles of that substance and vice versa; particles of a substance in a given mass of that …
• M1c Use ratios, fractions and percentages
• M2a Use an appropriate number of significant figures
• C1.3i explain how the mass of a given substance is related to the amount of that substance in moles and vice versa
• C1.3m deduce the stoichiometry of an equation from the masses of reactants and products and explain the effect of a limiting quantity of a reactant
• CM3.1i arithmetic computation and ratio when determining empirical formulae, balancing equations
• CM3.1iii provide answers to an appropriate number of significant figures
• WS.2b Make and record observations and measurements using a range of apparatus and methods
• C1.3h explain how the mass of a given substance is related to the amount of that substance in moles and vice versa
• C1.3k deduce the stoichiometry of an equation from the masses of reactants and products and explain the effect of a limiting quantity of a reactant
• C5.3.5 explain how the mass of a given substance is related to the amount of that substance in moles and vice versa and use the relationship: number of moles = mass of substance (g) / relative formula mass (g)
• C5.3.6 deduce the stoichiometry of an equation from the masses of reactants and products and explain the effect of a limiting quantity of a reactant
• IaS2.6 when processing data use an appropriate number of significant figures
• IaS2.11 in a given context interpret observations and other data (presented in diagrammatic, graphical, symbolic or numerical form) to make inferences and to draw reasoned conclusions, using appropriate scientific vocabulary and terminology to communicat…
• C5.2.5 explain how the mass of a given substance is related to the amount of that substance in moles and vice versa and use the relationship: number of moles = mass of substance (g) / relative formula mass (g)
• C5.2.6 deduce the stoichiometry of an equation from the masses of reactants and products and explain the effect of a limiting quantity of a reactant
• B2.19 The link between balanced equations and the ratio of moles of a substance in a reaction (for example, 2CH4 is 2 moles).
• In volatilisation conversion the substance is heated and any volatile products are driven off. It is heated to constant mass and the final mass recorded.

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

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The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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Bevans, R. (2023, June 22). Hypothesis Testing | A Step-by-Step Guide with Easy Examples. Scribbr. Retrieved June 9, 2024, from https://www.scribbr.com/statistics/hypothesis-testing/

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Experiment_602_Empirical Formula of MgO 1_4_2

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Experiment  60 2 :   Empirical Formula  

Section  1:  Purpose and Summary  

Determine the empirical formula of magnesium oxide. 

Calculate the mass of oxygen using weighing-by-difference. 

Calculate the mole of a sample from its mass. 

In this experiment, students will conduct the reaction between magnesium and oxygen gas.  Students will determine the mass of magnesium sample before and after the reaction, and the mass of magnesium and oxygen in the product.  Students will learn how to convert mass to mole of a given sample and determine empirical formula of a substance from mass and mole data. 

Section 2:  Safety Precautions and Waste Disposal  

Safety Precautions:  

Do not look directly at the burning magnesium ribbon. The flame is bright enough to damage your eye.  Use of eye protection is required for all experimental procedures.  

A hot crucible will break if placed directly on a cold surface.  Set hot crucibles on to wire screens to cool. 

A hot crucible will break if splashed with water directly.   Let crucibles cool prior to adding water.  

Waste Disposal:  

The solid product from the reaction can be disposed into the regular garbage can in the lab. 

Section 3: Procedure  

Part 1:   Preparation of the  c rucible  

Part  2 :    Preparation of  m agnesium  s ample  

Part 3:   Heating the  m agnesium  s ample  

Section  4:   Calculations    

Post Lab Questions:  

1.  There are some experimental errors that could lead to high or low mole ratio between Mg and O. In each case below, decide whether the situation described would lead to a calculated ratio of too much oxygen, or too little oxygen, and explain your answer. 

(a) You forgot to do the initial drying step and proceeded right away to weighing the crucible and lid you obtained from the stockroom. 

(b) Your magnesium ribbon is not shiny. But you did not polish it with steel wool prior to use as indicated in the experiment. 

(c) You added more laboratory water than is needed in Part 3 Step 7, and you did not dry it out completely. 

(d) After strong heating of the crucible you removed the lid but dropped it and broke. You then obtained a new lid for the final weighing. 

2. A similar experiment is performed to determine the empirical formula of an oxide of copper, and the following data were collected. Predict the empirical formula of the copper oxide from these data. 

Mass of crucible, cover, and copper sample 21.53 g 

Mass of empty crucible with cover 19.66 g 

Mass of crucible and cover and sample (after heating) 21.76 g 

How can the experiment for the determination of the empirical formula of an oxide of copper be improved?