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
