\$referrer_host = www.google.com. Percent Composition and Empirical Formula - Chemistry Steps

## Atoms, Molecules, and Ions

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 (H2SO4). If the element is not specified, we need to calculate the percentage of all the elements.

The molar mass of sulfuric acid is: M (H2SO4) = 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 (NH3).

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

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 C6H12O6 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:

C1H1.343O1

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

C3H4O3

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 C3H4O3, 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 is 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 C3H4O3 = C6H8O6

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

M (C6H8O6) = 6 x 12.0 + 8 x 1.00 + 6 x 16.0 = 176 g/mol

Check Also

#### Practice

1.

Calculate the percentage composition of butane, C4H10. How many percent of carbon and hydrogen, by mass, does butane contain?

82.8% C

17.2% H

Solution

2.

What is the percent composition of lithium in Li3PO?

17.9%

Solution

3.

What is the percent composition of water in Cupric sulfate pentahydrate, CuSO4·5H2O ?

36.1%

Solution

4.

How many grams of phosphorous are there in a 49.7 g sample of P2O5?

21.7 g of P

Solution

5.

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

CH4O3

Solution

6.

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 – C8H14O4

Empirical – C4H7O2

Solution

7.

What is the mass percent of the carbon in ethanol, C2H5OH?

52.2%

Solution

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 (C2H5OH) = 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:

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

8.

Calculate the percent composition by mass of the steroid hormone cortisol with the molecular formula of C21H30O5.

69.6% C, 8.29% H, 22.1% O

Solution

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

M (C21H30O5) = 21 x M (C) + 30 x M (H) + 5 x M (O) = 252 + 30 + 80 = 362 g/mol

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

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

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

9.

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

(a) Empirical formula CH2 (M = 84 g/mol)

(b) Empirical formula NH2 (M = 80 g/mol)

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

a) C6H12

b) N5H10

c) C4H4S4

Solution

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 CH2 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 (CH2) = C6H12

b) The molar mass of NH2 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 (NH2) = N5H10

c) The molar mass of NH2 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) = C4H4S4

10.

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.

N2O5

Solution

In chemical formulas, the subscripts indicate the moles of elements in one mole of the compound. So, for a preliminary formula, we can write N0.251O0.6273. 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 NO2.5. To simplify further, we multiply the numbers by two which allows having whole numbers. Therefore, the empirical formula of the compound is N2O5.

11.

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?

CCl4

Solution

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 C0.05147Cl0.2059. 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: CCl4.

12.

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.

C4H10POCl

Solution

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:

C2.85H7.17P0.7309O0.7144Cl0.7099

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

C4.01H10.10P0.7309O1.00Cl1

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

C4H10POCl

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

M (C4H10POCl) = 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 C4H10POCl.

13.

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 – C5H7N

molecular formula – C10H14N2

Solution

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:

C6.1667H8.7N1.2357

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

C4.99H7.04N

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

C5H7N

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 (C5H7N) = 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 (C5H7N) = C10H14N2

14.

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 – C3H7O2NS

Solution

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:

C2.4783H5.82O1.6506N0.8257S0.8246

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

C3.00H7.05O2NS

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

C3H7O2NS

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 (C3H7O2NS) = 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 – C3H7O2NS

15.

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 – CH2O

molecular formula – C3H6O3

Solution

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:

C3.3333H6.71O3.3313

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

CH2.01O1

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

CH2O

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 (CH2O) = 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 (CH2O) = C3H6O3

16.

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 CO2 and 0.4925 g of H2O were formed. Determine the molecular formula of acetone if its molas mass is 58.08 g/mol.

C3H6O

Solution

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

CHO + O2 → CO2 + H2O

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:

C0.01368H0.02736Ox

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:

C0.01368H0.02736O0.0045737

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

C2.991H5.982O1

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

C3H6O

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 (C3H6O) = 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 – C3H6O.

### 2 thoughts on “Percent Composition and Empirical Formula”

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

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