7.3 Percent composition and chemical formulas. Percent composition The relative amount of mass of each element in a compound, expressed in %

7.3 Percent composition and chemical formulas

Percent composition

• The relative amount of mass of each element in a compound, expressed in %

Calculating percent composition

% composition of an element = mass of element/mass of compound X 100

• The sum of all the % compositions of all the elements must equal 100%

Percent composition• We don’t know the mass of the element , but

we can find the percent composition using the formula and the molar mass of a substance

percent composition =

grams of element in 1 molemolar mass of compound

100

8.20 g Mg combines completely with 5.40 g O. What is the percent composition of

this compound?

8.20 g Mg combines completely with 5.40 g O. What is the percent

composition of this compound? • Add the masses of the elements• 8.20 g + 5.40 g = 13.60 g• Divide the mass of each element by the total

mass and multiply by 100 (make it a percent)• 8.20 g Mg / 13.60 g = 60.3 % Mg• That makes the % composition of O 39.7 %

222.6 g N combines completely with 77.4 g O. What is the percent composition of each element?

222.6 g N combines completely with 77.4 g O. What is the percent composition of each element?

• Total weight of compound is 300 g

• 222.6 g/300 g X 100 = 74.2 % N• 77.4 g/300 g X 100 = 25.8% O

• 74.2 %+ 25.8 % = 100%

Calculate the percent composition of HCN.

Calculate the percent composition of HCN.

• Molar mass H = 1 g• Molar mass C = 12 g• Molar mass N = 14 g• Total molar mass = 27 g• 1 g H/27 g X 100 = 3.7% H• 12 g C/27 g X 100 = 44.4% C• 14 g N/27 g X 100 = 51.9% N

Using the % composition of H in HCN (3.7%), calculate the amount of H in 378 g HCN.

Using the % composition of H in HCN (3.7%), calculate the amount of H in 378 g HCN.

• 378 g HCN X 3.7 g H/100 g HCN

378 g HCN 3.7 g H100 g HCN 14 g H

Conversion factor…There are 3.7 g of H in each 100 g HCN

Empirical formulas• Percent composition can be used to calculate

the empirical formula for a compound– Empirical formulas are the lowest whole

number ratios of the atoms in a compound– Empirical formulas may or may not be the

actual formula when we are dealing with molecules• For example, hydrogen peroxide, H2O2 has an

empirical formula of HO, but doesn’t occur in nature that way

Other cases where empirical formulas don’t tell us the composition of a molecule

• Empirical formula CH– Molecular formula C2H2 (ethyne or acetylene)

– Molecular formula C6H6 (benzene)

• Empirical formula CH2O– Molecular formula C2H4O2 (ethanoic acid)

– Molecular formula C6H12O6 (glucose)

What is the empirical formula of a compound which is 25.9% N and 74.1% O?

25.9 g N 1 mol N14 g N 1.85 mol N

74.1 g O 1 mol O16 g O 4.63 mol O

The ratio of moles of N to moles of O is 1.85/4.63, or 1/2.5. In whole numbers, this is a ratio of 2/5. Therefore, the empirical formula for this compound is N2O5.

Calculate the empirical formula of a compound which is 43.64% P and 56.36% O.

Calculate the empirical formula of a compound which is 43.64% P and 56.36% O.

43.64 g P 1 mol P30.97 g P 1.409 mol P

56.36 g O 1 mol O16 g O 3.523 mol O

The ratio of moles of P to moles of O is 1.409/3.523, or 1/2.5. In whole numbers, this is a ratio of 2/5. Therefore, the empirical formula for this compound is P2O5.

A compound is 2.477 g Mn and 1.323 g O. What is the empirical formula?

A compound is 2.477 g Mn and 1.323 g O. What is the empirical formula?

2.477 g Mn 1 mol Mn59.94 g Mn .0413 mol Mn

1.323 g O 1 mol O16 g O .0827 mol O

The ratio of moles of Mn to moles of O is .0413/.0827, or 1/2. Therefore, the empirical formula for this compound is MnO2.

Methyl butanoate has a percent composition of:58.8 % C; 9.8 % H; and 31.4 % O. Its molar mass is 102 grams. Want is its molecular formula?

Methyl butanoate has a percent composition of:58.8 % C; 9.8 % H; and 31.4 % O. Its molar mass is 102 grams. Want is its molecular formula?• .588 X 102 g = 59.98 g C• .098 X 102 g = 9.99 g H• .314 X 102 g = 32.03 g O These numbers look awfully suspicious! The first is 5 moles C, the second is 10 moles H,

and the third is 2 moles O. So the molecular formula is C5H10O2