A $10.5 \cdot g$ $10.5g$ mass of a hydrocarbon contains $1.5 \cdot g$ $1.5g$ of hydrogen. If the molecular mass of the hydrocarbon in $210 \cdot g \cdot m o l^{- 1}$ $210gmol^-1$ , what are the empirical and molecular formulae of the hydrocarbon?

Question

A $10.5 \cdot g$ $10.5g$ mass of a hydrocarbon contains $1.5 \cdot g$ $1.5g$ of hydrogen. If the molecular mass of the hydrocarbon in $210 \cdot g \cdot m o l^{- 1}$ $210gmol^-1$ , what are the empirical and molecular formulae of the hydrocarbon?

2 Answers

Impact of this question

2223 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-23T15:53:50

$1.5 g$ $1.5g$ of H would be $1.5 \div 1 = 1.5 m o l$ $1.5div1=1.5mol$
$9 g$ $9g$ of C would be $9 \div 12 = 0.75 m o l$ $9div12=0.75mol$

Explanation:

So the mol-ratio of $C \div H = 0.75 \div 1.5 = 1 \div 2$ $CdivH=0.75div1.5=1div2$
And the empirical formula is ${(C H_{2})}_{n}$ $(CH_2)_n$

One of those units has a mass of $12 + 2 \times 1 = 14 u$ $12+2xx1=14u$

So there are $210 \div 14 = 15$ $210div14=15$ of those units, and the molecular formula will be:

$C_{15} H_{30}$ $C_15H_30$

Answer 2 · 2017-04-23T16:02:21

We find $(i) the empirical formula$ $"(i) the empirical formula"$ , and then $(ii) the molecular formula$ $"(ii) the molecular formula"$ .

Explanation:

$Moles of hydrogen$ $"Moles of hydrogen"$ $=$ $=$ $\frac{1.5 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}} = 1.49 \cdot m o l$ $(1.5*g)/(1.00794*g*mol^-1)=1.49*mol$

$Moles of carbon$ $"Moles of carbon"$ $=$ $=$ $\frac{9.0 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 0.75 \cdot m o l$ $(9.0*g)/(12.011*g*mol^-1)=0.75*mol$ .

We divide thru by the smallest molar quantity, and (clearly) we get an empirical formula of $C H_{2}$ $CH_2$ .

But we know that the $molecular formula$ $"molecular formula"$ is whole number multiple of the $empirical formula$ $"empirical formula"$ ;

i.e. $molecular formula = n \times empirical formula$ $"molecular formula"=nxx"empirical formula"$

And thus, $210 \cdot g \cdot m o l = n \times (12.011 + 1.00794) \cdot g \cdot m o l^{- 1}$ $210*g*mol=nxx(12.011+1.00794)*g*mol^-1$ , and we solve for $n$ $n$ , to get $n = 16.15$ $n=16.15$ . This value should be closer to an integer, however, this gives a molecular formula of $C_{16} H_{32}$ $C_16H_32$ based on the figures you quoted.

Science

Math

Humanities

... and beyond

A $10.5 \cdot g$ $10.5g$ mass of a hydrocarbon contains $1.5 \cdot g$ $1.5g$ of hydrogen. If the molecular mass of the hydrocarbon in $210 \cdot g \cdot m o l^{- 1}$ $210gmol^-1$ , what are the empirical and molecular formulae of the hydrocarbon?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A 10.5*g10.5⋅g10.5*g mass of a hydrocarbon contains 1.5*g1.5⋅g1.5*g of hydrogen. If the molecular mass of the hydrocarbon in 210*g*mol^-1210⋅g⋅mol−1210*g*mol^-1, what are the empirical and molecular formulae of the hydrocarbon?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

A $10.5 \cdot g$ $10.5g$ mass of a hydrocarbon contains $1.5 \cdot g$ $1.5g$ of hydrogen. If the molecular mass of the hydrocarbon in $210 \cdot g \cdot m o l^{- 1}$ $210gmol^-1$ , what are the empirical and molecular formulae of the hydrocarbon?