The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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Answer:
10. Stacey
11. 7.86 g/cm^3 (3dp)
OR 786 kg/m^3 (SI units)
Explanation:
10. The correct exact answer for 10 is 26.169 (6.71x3.9). While Sam was correct in the precision of the number (2dp just like the given measurements), he was incorrect in the number itself. While Stacy's number was not as precise as Sam's (0dp), it is correctly rounded to the nearest whole number, therefore Stacy is correct.
11. The formula for density is mass/volume. The mass here is 264g and the volume is 33.6ml, therefore the density is 7.86 (3 sig figs because given values had 3) g/cm^3, because the mass was given in grams and 1ml = 1cm^3. Converted to SI units it is 786 kg/m^3.
Hope this helped!
Answer:
I think false idk if this is right
