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An alcohol is 62.04 % C and 10.43 % H by mass. The rest is oxygen. What is the empirical formula of the alcohol

User Nawin
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1 Answer

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Answer: The empirical formula is
C_3H_6O

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams. So, the mass of each element is equal to the percentage given.

Mass of C = 62.04 g

Mass of H = 10.43

Mass of O = (100-(62.04+10.43)) g = 27.53 g

Step 1 : convert given masses into moles.

Moles of C =
\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (62.04g)/(12g/mole)=5.17moles

Moles of H =
\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (10.43g)/(1g/mole)=10.43moles

Moles of O =
\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (27.53g)/(16g/mole)=1.72moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
(5.17)/(1.72)=3

For H =
(10.43)/(1.72)=6

For O =
(1.72)/(1.72)=1

The ratio of C :H : O= 3: 6 : 1

Hence the empirical formula is
C_3H_6O

User Stoneyan
by
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