Answer:
The answer to your question is: C₃H₃O This is my answer.
Explanation:
Data
Sample = 1.3109 g
CxHyOz
CO₂ = 3.2007 g
H₂O = 1.3102 g
Empirical formula = ?
MW CO2 = 44 g
MW H2O = 18 g
For Carbon
44 g -------------------- 12 g
3.2007 g ------------ x
x = (3.2007 x 12) / 44
x = 0.8729 g of Carbon
12 g of C -------------- 1 mol
0.8729 g -------------- x
x = (0.8729 x 1) / 12
x = 0.0727 mol of Carbon
For Hydrogen
18 g ---------------------- 1 g
1.3102 g ------------------- x
x = (1.3102 x 1) / 18
x = 0.0727 g of Hydrogen
1 g ------------------------ 1 mol
0.0727g ---------------- x
x = (0.0727 x 1)/1
x = 0.0727 mol of Hydrogen
For oxygen
g of Oxygen = g of sample - g of Carbon - g of hydrogen
g of Oxygen = 1.3109 - 0.8709 - 0.0727
g of Oxygen = 0.3673
16 g of Oxygen ------------- 1 mol of O
0.3673 g --------------------- x
x = (0.3673 x 1)/ 16
x = 0.0230 mol of Oxygen
Divide by the lowest number of moles
Carbon 0.0727 / 0.023 = 3.1 ≈ 3
Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3
Oxygen 0.0230 / 0.023 = 1
C₃H₃O
Answer:
The empirical formula of the compound is [tex]C_4H_8O_1[/tex]
Explanation:
Moles of carbon dioxide = [tex]\frac{3.2007 g}{44 g/mol}=0.07274 mol[/tex]
Moles of carbon in sample = [tex]1\times 0.07274 mol=0.07274 mol[/tex]
Moles of water = [tex]\frac{1.3101 g}{18 g/mol}=0.07279 mol[/tex]
Moles of hydrogen in sample = [tex]2\times 0.07279 mol=0.14558 mol[/tex]
Mass of sample = Mass of carbon + Mass of hydrogen + Mass of oxygen
Mass of oxygen =
1.3109 g - (12 g/mol × 0.07274 mol) - (1 g/mol× 0.14558 mol)
Mass of oxygen = 0.29244 g
Moles of oxygen in sample = [tex]\frac{0.29244 g}{16 g/mol}=0.01828 mol[/tex]
For empirical formula divide the moles of element which are is less amount with all the moles of every element.
Carbon =[tex]\frac{0.07274 mol}{0.01828 mol}=3.9\approx 4[/tex]
Hydrogen =[tex]\frac{0.14558 mol}{0.01828 mol}=7.9\approx 8[/tex]
oxygen=[tex]\frac{0.01828 mol}{0.01828 mol}=1[/tex]
The empirical formula of the compound is [tex]C_4H_8O_1[/tex]
