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Kazeer [188]
3 years ago
12

Combustion analysis of toluene, a common organic solvent, gives 4.69 mg of co2 and 1.10 mg of h2o. if the compound contains only

carbon and hydrogen, what is its empirical formula?
Chemistry
2 answers:
stich3 [128]3 years ago
8 0

Answer:

C_7H_8

Explanation:

Hello,

In this case, as toluene just have carbon and hydrogen, the moles coming from the 4.69mg of carbon dioxide match with the carbon moles contained into the toluene as well as for the hydrogen but for the 1.10mg of water which has two hydrogens inside, as shown below:

n_C=4.69mgCO_2*\frac{1mmolCO_2}{44mgCO_2} *\frac{1mmolC}{1mmolCO_2} =0.1066mmolC\\n_H=1.10mgH_2O*\frac{1mmolH_2O}{18mgH_2O} *\frac{2mmolH}{1mmolH_2O} =0.1222mmolC

Now, we divide both carbon and hydrogen moles by the carbon moles to get the number relating them as follows:

C=\frac{0.1066}{0.1066}=1;H=\frac{0.1222}{0.1066} =1.15

Finally, the factor multiplying such relations suitable to get whole numbers turns out being seven (7), therefore, toluene's empirical formula is:

C_7H_8

Best regards.

BartSMP [9]3 years ago
4 0
CₓHₐ + (x+a/4)O₂ = xCO₂ + a/2H₂O

n(CO₂)=4.69 mg/44.01 mg/mmol = 0.1066 mmol

n(H₂O)=1.10 mg/18.02 mg/mmol = 0.06104 mmol

C : H = 0.1066 : 0.06104*2 = 0.1066 : 0.12208 = 1 : 1.1453 = 7 : 8

C₇H₈ + 9O₂ = 7CO₂ + 4H₂O

C₇H₈
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