In an experiment 3.10g of carbon, hydrogen and oxygen produced 4.4g of C02 and 2.7g of water on complete combustion. Determine the empirical formula of the compound
( C= 12, 0= 16, H =1)
Given compound with C, H, O => 4.4-g CO₂ + 2.7-g H₂O
Elemental %Composition of …
%C in CO₂ = (12/44)100% = 27.3% => Wt C from CO₂ = 27.3% of 4.4-g CO₂ = 1.206-g Carbon
%O in CO₂ = (32/44)100% = 72.7% => Wt O from CO₂= 72.7% of 4.4-g CO₂ = 3.199-g Oxygen
%O in H₂O = (16/18)100% = 89.0% => Wt O from H₂O = 89.0% of 2.7-g H₂O = 2.403-g Oxygen
%H in H₂O = (2/18)100% = 11.0% => Wt H from H₂O = 11.0% of 2.7-g H₂O = 0.300-g Hydrogen
=> Σ (masses of elements in compounds formed) = 7.103-g (which corresponds with Σ of compound masses given 4.4-g + 2.7-g).
Converting to %-per 100-wt …
%C = (1.206/7.103)100% = 16.9%
%H = (0.300/7.103)100% = 4.2%
%O =[(3.199-g + 2.403-g)/7.103-g]100% = 78.9%
%C = 16.9% =>16.9-g per 100wt/12-g∙moleˉ¹ = 1.408-mole
%H = 4.2% => 4.2-g per 100wt/1-g∙moleˉ¹ = 4.200-mole
%0 = 78.9% => 78.9-g per 100wt/16-g∙moleˉ¹ = 4.931-mole
%Σ = 100% => Σ = 100-g sample
Empirical Ratio is then reduced mole ratios by dividing by the smaller of the mole values.
C:H:O => (1.408/1.408):(4.200/1.408):(4.931/1.403) => 1:3:3.5 x 2 => 2:6:7
Empirical formula => C₂H₆O₇
Note: After converting given data to grams per 100wt, calculation follows the progression …
% per 100wt => grams per 100wt => moles each element => reduce mole values by dividing by smallest mole value => Empirical Ratio => Empirical Formula
Good attempt
Good approach
Well, let's do some math tricks and unravel this empirical formula mystery, shall we?
First, we need to calculate the number of moles of carbon dioxide and water produced.
Moles of CO2 = mass of CO2 / molar mass of CO2
= 4.4g / (12g/mol + 2(16g/mol))
= 4.4g / (44g/mol)
= 0.1 mol
Moles of H2O = mass of H2O / molar mass of H2O
= 2.7g / (2(1g/mol) + 16g/mol))
= 2.7g / (18g/mol)
= 0.15 mol
Now, let's look at the mole ratios of carbon, hydrogen, and oxygen in the compound. It's like a three-way dance-off of elements!
The mole ratio of carbon to oxygen can be found by dividing the number of moles of carbon dioxide by the mole of carbon:
C:O = 0.1 mol C / 0.1 mol CO2 = 1 mol C / 1 mol CO2 = 1:1
Similarly, the mole ratio of hydrogen to oxygen can be found by dividing the number of moles of water by the mole of hydrogen:
H:O = 0.15 mol H / 0.15 mol H2O = 2 mol H / 1 mol H2O = 2:1
Now, let's combine these ratios to find the empirical formula!
So, the empirical formula of the compound is CH2O. Ta-da! We did it!
To determine the empirical formula of a compound, we need to calculate the ratio of the elements present in the compound.
1. Start by calculating the moles of carbon dioxide (CO2) and water (H2O) produced in the experiment.
- Molar mass of CO2 = 12 (C) + 2 * 16 (O) = 44 g/mol
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 4.4 g / 44 g/mol = 0.1 mol
- Molar mass of H2O = 2 (H) + 16 (O) = 18 g/mol
- Moles of H2O = mass of H2O / molar mass of H2O = 2.7 g / 18 g/mol = 0.15 mol
2. Next, calculate the moles of carbon (C) and hydrogen (H) present in the experiment.
- Moles of C = moles of CO2 = 0.1 mol
- Moles of H = 2 * moles of H2O = 2 * 0.15 mol = 0.3 mol
3. Finally, calculate the mole ratios of C, H, and O by dividing each by the smallest number of moles.
- Mole ratio of C = 0.1 mol / 0.1 mol = 1
- Mole ratio of H = 0.3 mol / 0.1 mol = 3
- Mole ratio of O = 0 (since oxygen only appears in CO2 and H2O)
Therefore, the empirical formula of the compound is CH3O.