Complete combustion of 0.327g of a compound contained H, C, N, O and S gave 0.55g of CO2, 0.081g of H2O and 0.02g of N, if 0.673g of the compound was converted to 0.859g of BeS. Calculate the empirical formular of the compound.
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To determine the empirical formula of the compound, we need to find the mole ratios of the elements present in the compound. Here's how we can calculate it step by step:
Step 1: Calculate the number of moles of each element:
Given:
- Mass of CO2 produced = 0.55 g
- Mass of H2O produced = 0.081 g
- Mass of N produced = 0.02 g
To calculate the number of moles, we can use the formula:
moles = mass / molar mass
The molar mass of CO2 (carbon dioxide) is:
C = 12.01 g/mol
O = 16.00 g/mol (x2 since there are two oxygen atoms)
Molar mass of CO2 = 12.01 + (16.00 x 2) = 44.01 g/mol
Number of moles of CO2 = 0.55 g / 44.01 g/mol = 0.0125 mol
Similarly, for H2O and N:
Molar mass of H2O (water) = 1.01 (2 hydrogen atoms) + 16.00 (1 oxygen atom) = 18.02 g/mol
Number of moles of H2O = 0.081 g / 18.02 g/mol = 0.0045 mol
Molar mass of N = 14.01 g/mol
Number of moles of N = 0.02 g / 14.01 g/mol = 0.0014 mol
Step 2: Determine the mole ratio of the elements:
Divide the number of moles of each element by the smallest mole value obtained in Step 1.
Dividing by 0.0014 (the smallest value):
CO2: 0.0125 mol / 0.0014 mol = 8.93
H2O: 0.0045 mol / 0.0014 mol = 3.21
N: 0.0014 mol / 0.0014 mol = 1
Step 3: Round the mole ratios to the nearest whole number:
To simplify the mole ratios, round each ratio to the nearest whole number. In this case, we get:
CO2: 9
H2O: 3
N: 1
The empirical formula is the simplest whole number ratio of atoms present in the compound. Therefore, the empirical formula of the compound is C9H3N.