Question

A compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this compound?

Answer 1

The solution would be like this for this specific problem:

 

Moles of carbon = 58.8 / 12 = 4.9 
Moles of hydrogen = 9.8 / 1 = 9.8 
Moles of oxugen = 31.4 / 16 m= 1.96 
Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1 

Simplest formula = C5H10


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Answer 2

Answer: Thus the molecular formula is $C_{5}H_{10}O_2$  

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 = 58.8 g

Mass of H = 9.8 g

Mass of O = 31.4 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{58.8g}{12g/mole}=4.9moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.8g}{1g/mole}=9.8moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{31.4g}{16g/mole}=1.96moles$

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

For C = $\frac{4.9}{1.96}=2.5$

For H = $\frac{9.8}{1.96}=5$

For O =$\frac{1.96}{1.96}=1$

The ratio of C : H : O= 2.5 : 5 : 1

Converting them into whole number ratios:

Hence the empirical formula is $C_{5}H_{10}O_2$  

The empirical weight of$C_{5}H_{10}O_2$  = 5(12) + 10(1) + 2(16)= 102g.

Given :The molecular weight = 102 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}$

$n=\frac{102g/mole}{102g/eq}=1$

Thus the molecular formula is $C_{5}H_{10}O_2$