The mass of the substance is 1.71 g. The mass of carbon dioxide is 2.64 g., The mass of water is 0.99 g

The mass of the substance is 1.71 g. The mass of carbon dioxide is 2.64 g., The mass of water is 0.99 g., Mr (substance) = 343 g / mol. Find the molecular formula (СxHyOz).

The reaction equation is:
X + O2 → CO2 ↑ + H2O.
Substance X contains C, H and O atoms.
Let’s find the amount of C, H and O, for this we calculate the amount of CO2 and H2O:
ν (CO2) = m (CO2) / M (CO2) = 2.64 / (12 + 16 * 2) = 0.06 mol.
ν (H2O) = m (H2O) / M (H2O) = 0.99 / 18 = 0.055 mol.
There is 1 C atom in 1 CO2 molecule (ratio 1: 1) ⇒ In 0.06 mol of CO2 0.06 mol of C atoms.
In 1 H2O molecule there are 2 H atoms (ratio 1: 2) ⇒ B 0.055 mol H2O 0.055 ∙ 2 = 0.11 mol of H atoms.
Let’s find the amount of O in the original substance:
m (substance) = m (C) + m (H) + m (O) ⇒ m (O) = m (substance) – m (C) – m (H).
m (C) = ν * M = 0.06 * 12 = 0.72 g.
m (H) = ν * M = 0.11 * 1 = 0.11 g.
m (O) = m (X) – m (C) – m (H).
m (O) = 1.71 – 0.72 – 0.11 = 0.88 g.
Determine the number of O atoms:
ν = m (O) / M (O) = 0.88 / 16 = 0.055 mol.
The ratio of C, H and O atoms in the formula of a substance: n (C): n (H): n (O) = 0.06: 0.11: 0.055 =
1.09: 2: 1.
We get the simple formula C1.09H2O1.
M (C1H2O1) = 12 * 1.09 + 1 * 2 + 16 = 31 g / mol.
The ratio M (X) / M (C1H2O1) = 343/31 = 11.
those. molecular formula X is a simple formula multiplied by 11 = (C1H2O1) 11 = C12H22O11.
Answer: the formula of the substance С12Н22О11 is sucrose.