barry um has a sample of a compound which weighs 200g and contains carbon,hydrogen,oxygen and nitrogen. by analysis, he find that it contains 97.56g of carbon,4.878g of hydrogen,52.03 of oxygen and 45.53g of nitrogen.find its empirical formula
To find the empirical formula, we need to determine the relative ratios of the elements present in the compound.
First, we find the moles of each element in the compound:
Moles of carbon = 97.56 g / molar mass of carbon
Moles of hydrogen = 4.878 g / molar mass of hydrogen
Moles of oxygen = 52.03 g / molar mass of oxygen
Moles of nitrogen = 45.53 g / molar mass of nitrogen
Next, we divide each mole value by the smallest mole value obtained. This will give us the mole ratios of the elements:
Moles of carbon (C) = Moles of carbon (97.56 g) / molar mass of carbon (12.01 g/mol) = 8.127 mol
Moles of hydrogen (H) = Moles of hydrogen (4.878 g) / molar mass of hydrogen (1.008 g/mol) = 4.845 mol
Moles of oxygen (O) = Moles of oxygen (52.03 g) / molar mass of oxygen (16.00 g/mol) = 3.251 mol
Moles of nitrogen (N) = Moles of nitrogen (45.53 g) / molar mass of nitrogen (14.01 g/mol) = 3.251 mol
Next, we convert the mole ratios to the nearest whole numbers by dividing each by the smallest value (3.251 mol):
C: 8.127 mol / 3.251 mol ≈ 2.5
H: 4.845 mol / 3.251 mol ≈ 1.5
O: 3.251 mol / 3.251 mol = 1
N: 3.251 mol / 3.251 mol = 1
Finally, we multiply each element by 2 to get whole numbers, since we obtained 2.5 for carbon and 1.5 for hydrogen:
C: 2 * 2.5 = 5
H: 2 * 1.5 = 3
O: 2
N: 2
Therefore, the empirical formula of the compound is C5H3O2N2.