Percentage composition shows the amount of each element in a compound expressed as a mass percent.

To calculate the percentage of an element, multiply its molar mass and the subscript which shows how many moles of it are present in one mole of the compound, then divide the product by the molar mass of the compound.

For example, let’s determine the percent composition of sulfuric acid (H_{2}SO_{4}). If the element is not specified, we need to calculate the percentage of all the elements.

The molar mass of sulfuric acid is: M (H_{2}SO_{4}) = 2 x 1.0 + 32 + 4 x 16.0 = 98 g/mol.

\[{\rm{\% }}\;{\rm{(H)}}\;{\rm{ = }}\;\frac{{{\rm{2}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(H)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{2}}\,{\rm{ \times }}\,{\rm{1}}{\rm{.0}}\,}}{{{\rm{98}}\,}}\,{\rm{ \times }}\,100\% \; = \;2.0\% \]

\[{\rm{\% }}\;{\rm{(S)}}\;{\rm{ = }}\;\frac{{{\rm{M}}\,{\rm{(S)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{32}}{\rm{.1}}\,}}{{{\rm{98}}\,}}\,{\rm{ \times }}\,100\% \; = \;33\% \]

\[{\rm{\% }}\;{\rm{(O)}}\;{\rm{ = }}\;\frac{{{\rm{4}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(O)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{4}}\,{\rm{ \times }}\,{\rm{16}}\,}}{{{\rm{98}}\,}}\,{\rm{ \times }}\,100\% \; = \;65\% \]

The sum of the percentages of all elements in the compound is 100%, so you can find the last one by subtracting the sum of the other elements from 100. This relies on the accuracy of the previous calculations, so double-check your calculation!

For example, let’s determine the percent composition of ammonia (NH_{3}).

\[{\rm{\% }}\;{\rm{(H)}}\;{\rm{ = }}\;\frac{{{\rm{3}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(H)}}}}{{{\rm{M}}\;{\rm{(N}}{{\rm{H}}_{\rm{3}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{3}}\,{\rm{ \times }}\,{\rm{1}}{\rm{.0}}\,}}{{{\rm{17}}\,}}\,{\rm{ \times }}\,100\% \; = \;17.6\% \]

Therefore, nitrogen makes 100 – 17.6 = 82.4% of ammonia by mass.

**Empirical Formula**

The empirical formula of a compound indicates the mole ratio of atoms in the simplest whole numbers. For example:

Notice that the empirical formula can be the same as the molecular formula if the numbers cannot be simplified any further.

**Determining the Empirical and Molecular Formulas**

To determine the empirical and molecular formulas, given the mass or percentage of the elements, you need to convert them to moles. This is because, remember, the formula of a compound tells us how many moles of each element there are in one mole of that compound.

For example, the formula of fructose C_{6}H_{12}O_{6 }indicates that in one mole of fructose, there are six moles of carbon, twelve moles of hydrogen, and six moles of oxygen.

Therefore, if you are asked to determine the empirical or molecular formula of a compound, you need to find the moles of each element.

**For example,** What are the empirical and molecular formulas of ascorbic acid if it consists of 40.92% carbon (C), 4.58% hydrogen (H), and 54.50% oxygen (O) and the molar mass is 176.1 g/mol?

The quantities are given in percentage, so we’ll go ahead and write these numbers as subscripts in the formula and convert them to moles following the steps in this plan:

**1)** The first step is converting the % to mass, and you may be wondering how to achieve this transformation. Remember, whenever the quantities are not given or not relatable, **you can assume a certain amount of the compound** because whether it is, for example, 100 g or 450 g of water, the formula is always the same.

Now, when going from % to mass, the easiest is to assume a 100 g sample since this converts to units from % to grams without changing the numbers. So, **assuming a 100 g sample**, we have 40.92% g C, 4.58% g H, and 54.50% g O.

**2)** Find the moles of C, H, and O using their molar masses:

\[{\rm{n}}\;{\rm{(C)}}\;{\rm{ = }}\,\frac{{{\rm{m}}\;}}{{{\rm{M}}\;}}\,{\rm{ = }}\;\frac{{{\rm{40}}{\rm{.92}}\;{\rm{g}}\;}}{{{\rm{12}}{\rm{.0}}\;{\rm{g/mol}}\;}}\;{\rm{ = }}\,{\rm{3}}{\rm{.41}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(H)}}\;{\rm{ = }}\,\frac{{{\rm{m}}\;}}{{{\rm{M}}\;}}\,{\rm{ = }}\;\frac{{{\rm{4}}{\rm{.58}}\;{\rm{g}}\;}}{{{\rm{1}}{\rm{.00}}\;{\rm{g/mol}}\;}}\;{\rm{ = }}\,{\rm{4}}{\rm{.58}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(C)}}\;{\rm{ = }}\,\frac{{{\rm{m}}\;}}{{{\rm{M}}\;}}\,{\rm{ = }}\;\frac{{{\rm{54}}{\rm{.50}}\;{\rm{g}}\;}}{{{\rm{16}}{\rm{.0}}\;{\rm{g/mol}}\;}}\;{\rm{ = }}\,{\rm{3}}{\rm{.41}}\;{\rm{mol}}\]

Don’t worry about significant figures here and keep at least three decimals.

**3)** Dividing the numbers by the smallest one is usually a good start to simplifying them. So, dividing by 3.41, the moles are:

**C _{1}H_{1.343}O_{1}**

_{ }We still need to simplify the moles of H. Remember, you can divide or multiply by any number to get a set of the simplest integers. Multiplying 1.343 by 3 is 4.029, which is approximately equal to 4 (4.029 ≈ 4). Make sure to multiply all the subscripts – you cannot multiply only one and leave the others because that would change the mole ratio of the elements. Therefore, by multiplying all the subscripts by 3, we get that the **empirical formula is **

**C _{3}H_{4}O_{3}**

**4)** Next, we need to check if the empirical formula is the same as the molecular formula of the compound. For this, **calculate the molar mass of the empirical formula** C_{3}H_{4}O_{3}, to see if it matches the actual molar mass given in the problem (176.1 g/mol):

3 x 12.0 + 4 x 1.00 + 3 x 16.0 = 88 g/mol

This indicates that the empirical formula of this compound i**s not the same as its molecular formula.**

Divide 176.1 by 88 to see how many times larger the molar mass of the molecular formula is compared to the one of the empirical formula:

176.1/88 ≈ 2

Therefore, the subscripts (moles) in the empirical formula must be multiplied by two to obtain the molecular formula:

**molecular formula = 2 x empirical formula**

2 x C_{3}H_{4}O_{3} = **C _{6}H_{8}O_{6}**

Calculate the molar mass of this formula to make sure it matches the one given in the problem:

M (C_{6}H_{8}O_{6}) = 6 x 12.0 + 8 x 1.00 + 6 x 16.0 = 176 g/mol

**Check Also**

- The Mole and Molar Mass
- Molar Calculations
- Percent Composition and Empirical Formula
- Stoichiometry of Chemical Reactions
- Limiting Reactant
- Reaction/Percent Yield
**Stoichiometry Practice Problems**

#### Practice

Calculate the percentage composition of butane, C_{4}H_{10}. How many percent of carbon and hydrogen, by mass, does butane contain?

82.8% C

17.2% H

What is the percent composition of lithium in Li_{3}PO_{4 }?

17.9%

What is the percent composition of water in Cupric sulfate pentahydrate, CuSO_{4}·5H_{2}O ?

36.1%

How many grams of phosphorous are there in a 49.7 g sample of P_{2}O_{5}?

21.7 g of P

Determine the empirical formula for a compound that contains 18.75 % C, 6.25% H, and 75% O by mass.

CH_{4}O_{3}

Fructone is used in different types of perfumes as a synthetic fruit-aroma additive. It contains 55.17 % C, 8.046 % H and the rest being oxygen. Determine the empirical and molecular formula of Fructone considering that its molar mass is 174.0 g/mol.

Molecular – C_{8}H_{14}O_{4}

Empirical – C_{4}H_{7}O_{2}

What is the mass percent of the carbon in ethanol, C_{2}H_{5}OH?

52.2%

The mass percent is calculated by the following formula:

Where n is the moles (subscript) of the element in the formula. We have the moles of carbon, so the missing part is the molar mass of ethanol:

M (C_{2}H_{5}OH) = 2 x M (C) + 6 x M (H) + 1 x M (O) = 2 x 12 + 6 x 1 + 1 x 16 = 46 g/mol

Therefore, for the % carbon, we can write:

\[{\rm{\% }}\;{\rm{(C)}}\;{\rm{ = }}\;\frac{{{\rm{2}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(C)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH)}}}}\;{\rm{ = }}\;\frac{{{\rm{2}}\,{\rm{ \times }}\,{\rm{12}}\,}}{{{\rm{46}}\,}}\,{\rm{ \times }}\,{\rm{100\% }}\;{\rm{ = }}\;{\rm{52}}{\rm{.2\% }}\]

Calculate the percent composition by mass of the steroid hormone *cortisol* with the molecular formula of C_{21}H_{30}O_{5}.

69.6% C, 8.29% H, 22.1% O

If a specific element is not mentioned, then you need to calculate the mass % of all. First, the molar mass of cortisol:

M (C_{21}H_{30}O_{5}) = 21 x M (C) + 30 x M (H) + 5 x M (O) = 252 + 30 + 80 = 362 g/mol

\[{\rm{\% }}\;{\rm{(C)}}\;{\rm{ = }}\;\frac{{{\rm{21}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(C)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{C}}_{{\rm{21}}}}{{\rm{H}}_{{\rm{30}}}}{{\rm{O}}_{\rm{5}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{21}}\,{\rm{ \times }}\,{\rm{12}}\,}}{{{\rm{362}}}}\,{\rm{ \times }}\,{\rm{100\% }}\;{\rm{ = }}\;{\rm{69}}{\rm{.6\% }}\]

\[{\rm{\% }}\;{\rm{(H)}}\;{\rm{ = }}\;\frac{{{\rm{30}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(H)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{C}}_{{\rm{21}}}}{{\rm{H}}_{{\rm{30}}}}{{\rm{O}}_{\rm{5}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{30}}\,{\rm{ \times }}\,{\rm{1}}\,}}{{{\rm{362}}}}\,{\rm{ \times }}\,{\rm{100\% }}\;{\rm{ = }}\;{\rm{8}}{\rm{.29\% }}\]

\[{\rm{\% }}\;{\rm{(O)}}\;{\rm{ = }}\;\frac{{{\rm{5}}\;{\rm{ \times }}\,{\rm{M}}\,{\rm{(O)}}}}{{{\rm{M}}\;{\rm{(}}{{\rm{C}}_{{\rm{21}}}}{{\rm{H}}_{{\rm{30}}}}{{\rm{O}}_{\rm{5}}}{\rm{)}}}}\;{\rm{ = }}\;\frac{{{\rm{5}}\,{\rm{ \times }}\,{\rm{16}}\,}}{{{\rm{362}}}}\,{\rm{ \times }}\,{\rm{100\% }}\;{\rm{ = }}\;{\rm{22}}{\rm{.1\% }}\]

Determine the molecular formula of each compound given the empirical formula.

(a) Empirical formula CH_{2} (M = 84 g/mol)

(b) Empirical formula NH_{2} (M = 80 g/mol)

(c) Empirical formula CHS (M = 180.4 g/mol)

a) C_{6}H_{12}

b) N_{5}H_{10}

c) C_{4}H_{4}S_{4}

To check if the empirical formula is the same as the molecular formula, or how many times smaller it is, we **calculate the molar mass of the empirical formula** and compare it with the molar mass of the molecular formula.

a) The molar mass of CH_{2 }is 14 g/mol. The ratio then would be:

\[\frac{{{\rm{M}}\;{\rm{(molecular)}}}}{{{\rm{M}}\;{\rm{(empirical)}}}}\; = \;\frac{{84}}{{14}}\; = \,6\]

Therefore, the molecular formula is:

6 x (CH_{2}) = **C _{6}H_{12}**

_{ }

b) The molar mass of NH_{2 }is 16 g/mol. The ratio then would be:

\[\frac{{{\rm{M}}\;{\rm{(molecular)}}}}{{{\rm{M}}\;{\rm{(empirical)}}}}\; = \;\frac{{80}}{{16}}\; = \,5\]

Therefore, the molecular formula is:

5 x (NH_{2}) = N_{5}H_{10}

c) The molar mass of NH_{2 }is 16 g/mol. The ratio then would be:

\[\frac{{{\rm{M}}\;{\rm{(molecular)}}}}{{{\rm{M}}\;{\rm{(empirical)}}}}\; = \;\frac{{180.4}}{{45.1}}\; = \,4\]

Therefore, the molecular formula is:

4 x (CHS) = C_{4}H_{4}S_{4}

Determine the empirical formula of the compound that is formed when 0.251 mol of nitrogen atoms are combined with 0.6273 mol of oxygen atoms.

N_{2}O_{5}

In chemical formulas, the subscripts indicate the moles of elements in one mole of the compound. So, for a preliminary formula, we can write N_{0.251}O_{0.6273. }To obtain the empirical formula, we can divide the subscripts by the smallest number, which is 0.251 in this case.

n (N) = 0.251 ÷ 0.251 = 1 mol

n (O) = 0.6273 ÷ 0.251 = 2.499 ≈ 2.5 mol

At this point, we have NO_{2.5}. To simplify further, we multiply the numbers by two which allows having whole numbers. Therefore, the **empirical formula** of the compound is **N _{2}O_{5}**

_{.}

In a reaction, 0.61764 g of carbon was combined with 7.309 g of chlorine. What is the empirical formula of the resulting product?

CCl_{4}

In this problem, we have the masses instead of the moles, so we first need to demine the moles of the elements.

\[{\rm{n}}\,{\rm{(C)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.61764}}\,\cancel{{\rm{g}}}\, \times \;\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{12}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.05147}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(Cl)}}\,{\rm{ = }}\;7.309\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{35}}{\rm{.5}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.2059}}\,{\rm{mol}}\,\]

For a preliminary formula, we can write C_{0.05147}Cl_{0.2059. }To obtain the empirical formula, we can divide the subscripts by the smallest number, which is 0.05147 in this case.

n (C) = 0.05147 ÷ 0.05147 = 1 mol

n (Cl) = 0.2059 ÷ 0.05147 = 4.00

We have obtained whole numbers, and therefore, these will be the moles of the elements in the empirical formula: **CCl _{4}.**

Determine the molecular formula of a compound that contains 34.2% C, 11.43% O, 22.0% P, 7.17% H, and 25.2% Cl, and has a molar mass of 140.55 g/mol.

C_{4}H_{10}POCl

If the masses of elements are not given, you can assume a 100 g sample. This allows switching from percentage to mass without any calculations. Therefore, we can say that the sample contains 34.2 g C, 22.0 g P, 7.17 g H, and 25.2 g Cl. The next step is to calculate the moles of the elements.

\[{\rm{n}}\,{\rm{(C)}}\,{\rm{ = }}\;34.2\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{12}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{2}}{\rm{.85}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(O)}}\,{\rm{ = }}\;11.43\,{\rm{g}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{16}}\,{\rm{g}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.7144}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(P)}}\,{\rm{ = }}\;22.0\,{\rm{g}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{30}}{\rm{.1}}\,{\rm{g}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.7309}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(H)}}\,{\rm{ = }}\;7.17\,{\rm{g}}\, \times \;\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{1}}\,{\rm{g}}}}\,{\rm{ = }}\,{\rm{7}}{\rm{.17}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(Cl)}}\,{\rm{ = }}\;25.2\,{\rm{g}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{35}}{\rm{.5}}\,{\rm{g}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.7099}}\,{\rm{mol}}\,\]

For the preliminary formula, we can write:

C_{2.85}H_{7.17}P_{0.7309}O_{0.7144}Cl_{0.7099}

To simplify, we divide the moles by the smallest number 0.7099:

C_{4.01}H_{10.10}P_{0.7309}O_{1.00}Cl_{1}

Rounding off the number, we get the following formula for the compound:

C_{4}H_{10}POCl

Next, check if the molar mass of this formula matches what is given in the problem:

M (C_{4}H_{10}POCl) = 4 x M (C) + 10 x M (H) + M (P) + M (O) + M (Cl) = 4 x 12 + 10 x 1 + 31 + 16 + 35.5 = 140.5 g/mol

The molar mass is the same as given in the problem, therefore, the molecular formula of the compound is **C _{4}H_{10}POCl.**

Nicotine contains 74.0% carbon, 8.7% hydrogen, and 17.3% nitrogen by mass. What are the empirical and molecular formulas of nicotine considering its molar mass is 162.1 g/mol.

empirical formula – C_{5}H_{7}N

molecular formula – C_{10}H_{14}N_{2}

If the masses of elements are not given, you can assume a 100 g sample. This allows switching from percentage to mass without any calculations. Therefore, we can say that the sample contains 74.0 g C, 8.7 g H, and 17.3 g N. The next step is to calculate the moles of the elements.

\[{\rm{n}}\,{\rm{(C)}}\,{\rm{ = }}\;74.0\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{12}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{6}}{\rm{.1667}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(H)}}\,{\rm{ = }}\;8.7\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{1}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{8}}{\rm{.7}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(N)}}\,{\rm{ = }}\;17.3\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{14}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{1}}{\rm{.2357}}\,{\rm{mol}}\,\]

For the preliminary formula, we can write:

C_{6.1667}H_{8.7}N_{1.2357}

To simplify, we divide the moles by the smallest number 1.2357:

C_{4.99}H_{7.04}N

Rounding off the number, we get the following formula for the compound:

C_{5}H_{7}N

Now, **this is the empirical formula of nicotine** as it contains the smallest possible whole numbers given the molar ratio of the elements. Next, check if the molar mass of this formula matches what is given in the problem:

M (C_{5}H_{7}N) = 5 x M (C) + 7 x M (H) + M (N) = 5 x 12 + 7 x 1 + 14 = 81 g/mol

To check if the empirical formula is the same as the molecular formula, or how many times smaller it is, we **calculate the molar mass of the empirical formula** and compare it with the molar mass of the molecular formula.

The molar mass of nicotine is 162.1 g/mol. The ratio than would be:

\[\frac{{{\rm{M}}\;{\rm{(molecular)}}}}{{{\rm{M}}\;{\rm{(empirical)}}}}\; = \;\frac{{162.1}}{{81}}\; = \,2\]

The molar mass of nicotine is twice that of the empirical formula, therefore, to obtain the **molecular formula**, we must multiply all the subscripts by two:

2 x (C_{5}H_{7}N) = **C _{10}H_{14}N_{2}**

Cysteine is an important non-essential amino acid that contains 29.74 % carbon, 5.82 % hydrogen, 26.41 % oxygen,11.56 % nitrogen, and 26.47 % sulfur. What are the empirical and molecular formulas of cysteine considering it molar mass is 121.2 g/mol.

Both empirical and molecular formulas are identical –** C _{3}H_{7}O_{2}NS**

If the masses of elements are not given, you can assume a 100 g sample. This allows to switch from percentage to mass without any calculations. Therefore, we can say that the sample contains 29.74 g C, 5.82 g H, 26.41 g O, 11.56 g N, and 26.47 g S. The next step is to calculate the moles of the elements.

\[{\rm{n}}\,{\rm{(C)}}\,{\rm{ = }}\;29.74\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{12}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{2}}{\rm{.4783}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(H)}}\,{\rm{ = }}\;5.82\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{1}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{5}}{\rm{.82}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(O)}}\,{\rm{ = }}\;26.41\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{16}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{1}}{\rm{.6506}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(N)}}\,{\rm{ = }}\;11.56\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{14}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.8257}}\;{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(S)}}\,{\rm{ = }}\;26.47\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{32}}{\rm{.1}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.8246}}\;{\rm{mol}}\,\]

For the preliminary formula, we can write:

C_{2.4783}H_{5.82}O_{1.6506}N_{0.8257}S_{0.8246}

To simplify, we divide the moles by the smallest number 0.8246:

C_{3.00}H_{7.05}O_{2}NS

Rounding off the number, we get the following formula for the compound:

C_{3}H_{7}O_{2}NS

Now, **this is the empirical formula of cysteine** as it contains the smallest possible whole numbers given the molar ratio of the elements. Next, check if the molar mass of this formula matches what is given in the problem:

M (C_{3}H_{7}O_{2}NS) = 3 x M (C) + 7 x M (H) + 2 x M (O) + M (N) + M (S) = 3 x 12 + 7 x 1 + 2 x 16 + 14 + 32.1 = 121.1 g/mol

Because the molar mass of the molecular formula matches with the molar mass of cysteine given in the problem, the empirical and molecular formulas are identical – **C _{3}H_{7}O_{2}NS**

Muscle soreness during excessive physical activity is due to the synthesis of lactic acid that contains 40.0 % C, 6.71 % H, and 53.3 % O by mass. Determine the empirical and molecular formulas of lactic acid considering it molar mass is 90.08 g/mol.

empirical formula – CH_{2}O

molecular formula – C_{3}H_{6}O_{3}

If the masses of elements are not given, you can assume a 100 g sample. This allows switching from percentage to mass without any calculations. Therefore, we can say that the sample contains 40.0 g C, 6.71 g H, and 53.3 g O. The next step is to calculate the moles of the elements.

\[{\rm{n}}\,{\rm{(C)}}\,{\rm{ = }}\;40.0\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{12}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{3}}{\rm{.3333}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(H)}}\,{\rm{ = }}\;6.71\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{1}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{6}}{\rm{.71}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(O)}}\,{\rm{ = }}\;53.3\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{16}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{3}}{\rm{.3313}}\,{\rm{mol}}\,\]

For the preliminary formula, we can write:

C_{3.3333}H_{6.71}O_{3.3313}

To simplify, we divide the moles by the smallest number 3.3333:

CH_{2.01}O1

Rounding off the number, we get the following formula for the compound:

CH_{2}O

Now, **this is the empirical formula of lactic acid** as it contains the smallest possible whole numbers given the molar ratio of the elements. Next, check if the molar mass of this formula matches what is given in the problem:

M (CH_{2}O) = 30 g/mol

To check if the empirical formula is the same as the molecular formula, or how many times smaller it is, we **calculate the molar mass of the empirical formula** and compare it with the molar mass of the molecular formula.

The molar mass of lactic acid is 90.08 g/mol. The ratio then would be:

\[\frac{{{\rm{M}}\;{\rm{(molecular)}}}}{{{\rm{M}}\;{\rm{(empirical)}}}}\; = \;\frac{{90.08}}{{30}}\; = \,3\]

The molar mass of lactic acid is three times that of the empirical formula, therefore, to obtain the **molecular formula**, we must multiply all the subscripts by three:

3 x (CH_{2}O) = **C _{3}H_{6}O_{3}**

Acetone is commonly used as a nail polish remover. When 0.2647 g of acetone was burned in a combustion apparatus, 0.6019 g of CO_{2} and 0.4925 g of H_{2}O were formed. Determine the molecular formula of acetone if its molas mass is 58.08 g/mol.

C_{3}H_{6}O

We can write a general equation for the combustion of acetone considering it only contains C, H, and O atoms:

CHO + O_{2} → CO_{2} + H_{2}O

The plan is to determine the moles of C and O, and also the oxygen that is in the acetone. So, for this, convert the mass to moles:

\[{\rm{n}}\,{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\,{\rm{ = }}\;0.6019\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{44}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.01368}}\,{\rm{mol}}\,\]

\[{\rm{n}}\,{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;0.4925\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{18}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.02736}}\,{\rm{mol}}\,\]

At this point, we can write a preliminary formula for acetone based on the moles of C and H. Notice that the only source of C and H in CO2 and H2O is acetone, and therefore, according to the law of conservation of mass, acetone must have contained 0.01368 mol of carbon, and 0.02736 mol of hydrogen. So, the preliminary formula would be:

C_{0.01368}H_{0.02736}O_{x}

X is the moles of oxygen, and that is what we need to determine next. For this, we calculate the masses of carbon and hydrogen and subtract the sum from the mass of the acetone sample.

\[{\rm{m}}\,{\rm{(C)}}\,{\rm{ = }}\;0.01368\,\cancel{{{\rm{mol}}}}\, \times \,\frac{{{\rm{12}}\,{\rm{g}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.16416 g}}\,\]

\[{\rm{m}}\,{\rm{(H)}}\,{\rm{ = }}\;0.02736\,\cancel{{{\rm{mol}}}}\, \times \,\frac{{{\rm{1}}\,{\rm{g}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}}}}}\,{\rm{ = }}\,0.02736{\rm{ g}}\,\]

The sum of the masses would be 0.16416 + 0.02736 = 0.19152 g, and therefore, the mass of oxygen in the sample of acetone is:

m (O) = 0.2647 – 0.19152 = 0.07318 g

Next, we calculate the moles of oxygen to put it in the preliminary formula:

\[{\rm{n}}\,{\rm{(O)}}\,{\rm{ = }}\;0.07318\,\cancel{{\rm{g}}}\, \times \,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{16}}\,\cancel{{\rm{g}}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.0045737}}\,{\rm{mol}}\,\]

For the preliminary formula, we can write:

C_{0.01368}H_{0.02736}O_{0.0045737}

To simplify, we divide the moles by the smallest number 0.0045737:

C_{2.991}H_{5.982}O1

Rounding off the number, we get the following formula for the compound:

C_{3}H_{6}O

Now, **this is the empirical formula of acetone** as it contains the smallest possible whole numbers given the molar ratio of the elements. Next, check if the molar mass of this formula matches what is given in the problem:

M (C_{3}H_{6}O) = 58 g/mol

Because the molar mass of the molecular formula matches the molar mass of acetone given in the problem, the empirical and molecular formulas are identical – **C _{3}H_{6}O.**

Hi, I have a question. Why did you multiply 3 to 1.343 for Hydrogen? I am confused of where that 3 came from.

Hi,

So, the idea of getting the empirical formula is to divide or multiply the subscripts by any number to get whole numbers. The 3s of C and O are coming from multiplying their subscripts by 3 which makes the H equal to 4 since we have to multiply/divide the moles of all the elements in the formula.