A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is found to have a density of 1.87 g/L, and 1 mol of it occupies a volume of 22.4 L. Answer the following questions to determine two possible Lewis structures for molecules of the compound. What is the empirical formula of the compound?

The empirical formula of the compound is CH2

Explanation:

Step 1: Data given

A substance contains 85.7 % carbon and 14.3 % hydrogen.

The substance has a density of 1.87 g/L

1 mol occupies 22.4 L

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate molar mass of the substance

Since 1 mol occupies 22.4 L;

1 mol of this substance = 1.87g/L * 22.4 = 41.888 grams

This means the molar mass of the substance is 41.888 g/mol

Step 3: Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

Step 4: Calculate moles of carbon

moles C = mass C / Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

Step 5: Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

Step 6 : Calculate moles of hydrogen

Moles H = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

Step 7: Calculate mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

Step 8: calculate molar formule

Molar mass of empirical formule = 14.02 g/mol

n = Molar mass of substance / molar mass of empirical formule

n = 41.888 / 14.02 = 3

This means we have to multiply the empirical formula by 3

3 * (CH2) = C3H6

C3H6 can be propene or cyclopropane