A1.35g sample of asubstance containing carbon hydrogen nitrogen and oxcgen on combustion produced 0.810gof water and 1.32g of carbon dioxide In separate exp all the nitrogen in a0.735 g sample of the substance was converted into ammonia the mass of ammonia obtained was 0.284g determine empirical formulae of the substance
To determine the empirical formula of the substance, we need to find the mole ratio of the elements present in it. We can achieve this by following these steps:
Step 1: Calculate the moles of water (H2O) and carbon dioxide (CO2) produced.
To calculate the moles, we use the formula:
moles = mass / molar mass
The molar mass of water (H2O) = 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
moles of water = 0.810 g / 18.016 g/mol = 0.045 mol
The molar mass of carbon dioxide (CO2) = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
moles of carbon dioxide = 1.32 g / 44.01 g/mol = 0.030 mol
Step 2: Determine the moles of hydrogen (H), carbon (C), and oxygen (O).
In water, there are two moles of hydrogen for every one mole of water.
moles of hydrogen = 0.045 mol × 2 = 0.090 mol
In carbon dioxide, there is one mole of carbon for every one mole of carbon dioxide.
moles of carbon = 0.030 mol
In carbon dioxide, there are two moles of oxygen for every one mole of carbon dioxide.
moles of oxygen = 0.030 mol × 2 = 0.060 mol
Step 3: Calculate the moles of nitrogen (N) from the ammonia (NH3) obtained.
The molar mass of ammonia (NH3) = 14.01 g/mol + 3(1.008 g/mol) = 17.035 g/mol
moles of ammonia = 0.284 g / 17.035 g/mol = 0.0167 mol
Step 4: Compare the moles of nitrogen (N) with the moles of carbon (C), hydrogen (H), and oxygen (O).
To do this, divide each mole value by the smallest mole value (0.0167 mol).
moles of nitrogen (N) = 0.0167 mol / 0.0167 mol = 1
moles of hydrogen (H) = 0.090 mol / 0.0167 mol ≈ 5.39
moles of carbon (C) = 0.030 mol / 0.0167 mol ≈ 1.80
moles of oxygen (O) = 0.060 mol / 0.0167 mol ≈ 3.59
Step 5: Convert the mole ratios to whole numbers.
Since we want to determine the empirical formula, we need to find the simplest whole-number ratio. In this case, we can multiply all of the mole values by a factor to get the simplest ratios:
moles of nitrogen (N) = 1 × 1 = 1
moles of hydrogen (H) = 5.39 × 1 = 5
moles of carbon (C) = 1.80 × 1 = 2
moles of oxygen (O) = 3.59 × 1 = 4
The empirical formula of the substance is N1H5C2O4.