Question

An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If the compound has a molecular weight of 112 g/mol, what is its molecular formula?

Answer 1

Answer:

The molecular formula of this compound is C4H4N2O2

Explanation:

Step 1: Data given

Mass of the compound = 84 mg

The compound contains:

36 mg of Carbon

3 mg of hydrogen

21 mg of nitrogen

24 mg of oxygen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Molar mass of nitrogen = 14 g/mol

Molar mass of oxygen = 16 g/mol

Step 2: Calculate number of moles

Moles = mass / molar mass

Moles of carbon = 0.036 g/ 12.01 g/mol = 0.003 moles

Moles of hydrogen = 0.003 g/ 1.01 g/mol = 0.003 moles

Moles of nitrogen = 0.021 g/ 14 g/mol = 0.0015 moles

Moles of oxygen = 0.024 g/ 16 g/mol = 0.0015 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.003 / 0.0015 = 2

H: 0.003 / 0.0015 = 2

N = 0.0015/0.0015 = 1

O = 0.0015/0.0015 = 1

The empirical formula is C2H2NO

The molecular mass of this empirical formula is 56 g/mol

Step 4: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 112 g/mol / 56g/mol = 2

We have to multiply the empirical formula by 2

Molecular formula = 2*(C2H2NO) = C4H4N2O2

The molecular formula of this compound is C4H4N2O2