Determine the molecular formula of organic matter if, when burning 2.45 g of this substance, 7.7 g
Determine the molecular formula of organic matter if, when burning 2.45 g of this substance, 7.7 g of carbon dioxide and 3.15 g of water were formed. The relative vapor density of this substance in terms of carbon dioxide is 0.64
Since only carbon dioxide and water are formed during the combustion of a substance, it can include only three elements in its composition: carbon, hydrogen and oxygen.
Let’s find the molar mass of the substance:
M (substance) = M (CO2) * D CO2 (substance) = 44 * 0.64 = 28 g / mol.
n (substances) = 2.45 / 28 = 0.0875 mol;
Let’s find the amount of substance of carbon and hydrogen atoms in the specified amount of substance:
n (C) = n (CO2) = m (CO2) / M (CO2) = 7.7 / 44 = 0.175 mol;
n (H) = 2 * n (H2O) = 2 * (m (H2O) / M (H2O)) = 2 * (3.15 / 18) = 0.35 mol.
One molecule of a substance will contain:
0.175 / 0.0875 = 2 carbon atoms;
0.35 / 0.0875 = 4 hydrogen atoms;
Suppose there are no oxygen atoms in the substance. Then its molar mass will be:
M (substance) = 12 * 2 + 1 * 4 = 28 – coincides with the one found initially.
Substance: C2H4 – ethylene.
Answer: C2H4 is ethylene.