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kati45 [8]
3 years ago
14

A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is fou

nd 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?
Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer:

The empirical formula of the compound is CH2

Explanation:

<u>Step 1:</u> 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

<u>Step 2</u>: 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

<u>Step 3:</u> Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

<u>Step 4: </u>Calculate moles of carbon

moles C = mass C/ Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

<u>Step 5:</u> Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

<u>Step 6 :</u>Calculate moles of hydrogen

Moles H  = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

<u>Step 7: </u>Calculate  mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

<u>Step 8</u>: 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

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