$\mbox{RE}=\mbox{CH}_{2}\mbox{O}$

$\begin{aligned}M_{r}\mbox{RE} & =(1\times A_{r}\mbox{ C})+(2\times A_{r}\mbox{ H})+(1\times A_{r}\mbox{ O})\\
& =(1\times12)+(2\times1)+(1\times16)\\
& =12+2+16\\
& =30
\end{aligned}
$

$\begin{alignedat}{1}x & =\frac{M_{r}\mbox{ RM}}{M_{r}\mbox{ RE}}\\
& =\frac{180}{30}\\
& =6\\
\mbox{RM} & =x(\mbox{RE})\\
& =6(\mbox{CH}_{2}\mbox{O})\\
& =\mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6}.
\end{alignedat}
$

Perbandingan unsur penyusun suatu senyawa = rumus empiris.

$\mbox{massa unsur C}=7,2\mbox{ gram}$

$\mbox{massa unsur H}=1,2\mbox{ gram}$

massa unsur O $=$ massa senyawa $-$ (massa unsur C $+$ massa unsur H)

$\begin{aligned} & =18\mbox{ gram}-(7,2\mbox{ gram}+1,2\mbox{ gram})\\
& =9,6\mbox{ gram}.
\end{aligned}
$

Selanjutnya untuk mencari perbandingan unsur dalam senyawa diperoleh dengan cara berikut,

$\begin{array}{ccccc}
\mbox{mol C} & : & \mbox{mol H} & : & \mbox{mol O}\\
\frac{\mbox{massa C}}{A_{r}\mbox{ C}} & : & \frac{\mbox{massa H}}{A_{r}\mbox{ H}} & : & \frac{\mbox{massa O}}{A_{r}\mbox{ O}}\\
\frac{7,2}{12} & : & \frac{1,2}{1} & : & \frac{9,6}{16}\\
0,6 & : & 1,2 & : & 0,6\\
1 & : & 2 & : & 1.
\end{array}$

$M_{r}\mbox{ rumus molekul}=78$

$\begin{alignedat}{1}M_{r}\mbox{ rumus empiris} & =A_{r}\mbox{ C}+A_{r}\mbox{ H}\\
& =12+1\\
& =13
\end{alignedat}
$

$\begin{alignedat}{1}\mbox{Rumus molekul} & =x\times\mbox{rumus molekul}\\
x & =\frac{M_{r}\mbox{ rumus molekul}}{M_{r}\mbox{ rumus empiris}}\\
& =\frac{78}{13}\\
& =6\\
\mbox{Rumus molekul} & =6\times\mbox{CH}\\
& =\mbox{C}_{6}\mbox{H}_{6}.
\end{alignedat}
$

Rumus empiris adalah perbandingan paling sederhana dari unsur-unsur penyusun senyawa.

$\begin{aligned}\mbox{mol C} & :\mbox{ mol H}\\
\frac{\mbox{massa C}}{A_{r}\mbox{ C}} & :\frac{\mbox{massa H}}{A_{r}\mbox{ H}}\\
\frac{85,71}{12} & :\frac{14,29}{1}\\
7,1425 & :14,29
\end{aligned}
$

Hasil mol terkecil dijadikan sebagai pembagi untuk menentukan perbandingannya.

$\begin{aligned}\frac{7,1425}{7,1425} & :\frac{14,29}{7,1425}\\
1 & :\,2
\end{aligned}
$

Dari perhitungan diperoleh mol C : mol H =1 : 2, sehingga rumus empiris senyawa hidrokarbon tersebut adalah $\mbox{CH}_{2}$.

(a) Rumus empiris dari $\mbox{C}_{3}\mbox{H}_{8}$ sama dengan rumus molekulnya, karena perbandingan unsur penyusunnya tidak bisa disederhanakan lagi.

(b) Rumus empiris $\mbox{C}_{6}\mbox{H}_{10}=\mbox{C}_{3}\mbox{H}_{5}$ (masing-masing jumlah unsur dibagi 2).

$\begin{alignedat}{1}\mbox{C}_{3}\mbox{H}_{5} & =\mbox{C}_{3}\mbox{H}_{3+2}\\
& =\mbox{C}_{n}\mbox{H}_{n+2}
\end{alignedat}
$

(c) Rumus empiris $\mbox{C}_{8}\mbox{H}_{18}$=$\mbox{C}_{4}\mbox{H}_{9}$ (masing-masing jumlah unsur dibagi 2)

$\begin{alignedat}{1}\mbox{C}_{4}\mbox{H}_{9} & =\mbox{C}_{4}\mbox{H}_{4+5}\\
& =\mbox{C}_{n}\mbox{H}_{n+5}
\end{alignedat}
$

(d) Rumus empiris $\mbox{C}_{4}\mbox{H}_{10}=\mbox{C}_{2}\mbox{H}_{5}$ (masing-masing jumlah unsur dibagi 2)

$\begin{alignedat}{1}\mbox{C}_{2}\mbox{H}_{5} & =\mbox{C}_{2}\mbox{H}_{2+3}\\
& =\mbox{C}_{n}\mbox{H}_{n+3}
\end{alignedat}
$

(e) Rumus empiris $\mbox{C}_{6}\mbox{H}_{8}=\mbox{C}_{3}\mbox{H}_{4}$ (masing-masing jumlah unsur dibagi 2)

$\begin{alignedat}{1}\mbox{C}_{3}\mbox{H}_{4} & =\mbox{C}_{3}\mbox{H}_{3+1}\\
& =\mbox{C}_{n}\mbox{H}_{n+1}.
\end{alignedat}
$

$\underset{5\mbox{ gram}}{\mbox{C}_{x}\mbox{H}_{y}\mbox{O}_{z}}+\mbox{O}_{2}$$\rightarrow\underset{10\mbox{ gram}}{\mbox{CO}_{2}}+\underset{4\mbox{ gram}}{\mbox{H}_{2}\mbox{O}}$

Massa zat sebelum dan sesudah reaksi adalah sama, sehingga massa unsur karbon (C) dalam senyawa ester $\left(\mbox{C}_{x}\mbox{H}_{y}\mbox{O}_{z}\right)$ sama dengan massa unsur karbon (C) di $\mbox{CO}_{2}$. Kita dapat mencari massa karbon (C) dalam senyawa $\mbox{CO}_{2}$ dengan memakai kadar unsur dalam senyawa.

$\begin{aligned}\mbox{Massa C dalam CO}_{2} & =\frac{A_{r}\mbox{ C}\times\mbox{jumlah atom C}}{M_{r}\mbox{ CO}_{2}}\times\mbox{massa CO}_{2}\\
& =\frac{12\times1}{44}\times10\mbox{ gram}\\
& =2,727\mbox{ gram}
\end{aligned}
$

Dengan cara yang sama, kita dapat mencari massa H dalam $\mbox{H}_{2}\mbox{O}$

$\begin{aligned}\mbox{Massa H dalam H}_{2}\mbox{O} & =\frac{\mbox{massa H}\times\mbox{jumlah atom H}}{M_{r}\mbox{ H}_{2}\mbox{O}}\times\mbox{massa H}_{2}\mbox{O}\\
& =\frac{1\times2}{18}\times4\mbox{ gram}\\
& =0,444\mbox{ gram}
\end{aligned}
$.

Untuk mencari massa unsur oksigen (O) dapat diperoleh dari:

$\begin{aligned}\mbox{Massa C}_{x}\mbox{H}_{y}\mbox{O}_{z} & =\mbox{massa C+massa H+massa O}\\
5\mbox{ gram} & =2,727\mbox{ gram}+0,4444\mbox{ gram}+\mbox{massa O}\\
\mbox{massa O} & =5\mbox{ gram}-(2,727\mbox{ gram}+0,4444\mbox{ gram})\\
& =1,829\mbox{ gram}
\end{aligned}
$

Rumus empiris adalah perbandingan paling sederhana dari unsur-unsur penyusun senyawa.

$\begin{array}{ccccc}
\mbox{mol C} & : & \mbox{mol H} & : & \mbox{mol O}\\
\frac{\mbox{massa C}}{A_{r}\mbox{ C}} & : & \frac{\mbox{massa H}}{A_{r}\mbox{ H}} & : & \frac{\mbox{massa O}}{A_{r}\mbox{ O}}\\
\frac{2,727}{12} & : & \frac{0,444}{1} & : & \frac{1,829}{16}\\
0,227 & : & 0,444 & : & 0,114
\end{array}$

Hasil mol terkecil dijadikan sebagai pembagi untuk menentukan perbandingannya.

$\begin{array}{ccccc}
\frac{0,227}{0,114} & : & \frac{0,444}{0,114} & : & \frac{0,114}{0,114}\\
1,99 & : & 3,89 & : & 1\\
2 & : & 4 & : & 1
\end{array}$

Dari perhitungan diperoleh mol C : mol H : mol O = 2 : 4 : 1, sehingga rumus empiris senyawa ester tersebut adalah $\mbox{C}_{2}\mbox{H}_{4}\mbox{O}$.

Rumus molekul = $x$$\times$ rumus empiris ($x$ adalah faktor kali)

Untuk mencari nilai x digunakan persamaan berikut:

$\begin{alignedat}{1}x & =\frac{M_{r}\mbox{ rumus molekul}}{M_{r}\mbox{ rumus empiris}}\\
& =\frac{88}{44}\\
& =2
\end{alignedat}
$

Sehingga,

$\begin{alignedat}{1}\mbox{Rumus molekul} & =2\times\mbox{C}_{2}\mbox{H}_{4}\mbox{O}_{2}\\
& =\mbox{C}_{4}\mbox{H}_{8}\mbox{O}_{4}.
\end{alignedat}
$

$\begin{alignedat}{1}\mbox{massa L} & =\frac{52,94}{100}\times10\mbox{ gram}\\
& =5,294\mbox{ gram}\\
\mbox{massa O} & =10-5,294\mbox{gram}\\
& =4,706\mbox{ gram}
\end{alignedat}
$

$\begin{array}{ccccccc}
\mbox{L} & : & \mbox{O}\\
\frac{5,294}{A_{r}\mbox{ L}} & : & \frac{4,706}{16} & = & 2 & : & 3\\
\frac{5,294}{A_{r}\mbox{ L}} & : & 0,294 & = & 2 & : & 3\\
& & \frac{5,294}{A_{r}\mbox{ L}}\times3 & = & 0,588\\
& & A_{r}\mbox{ L} & = & 27.
\end{array}$

$\begin{array}{ccc}
\mbox{C}_{x}\mbox{H}_{2x+2}(g) & + & \mbox{O}_{2}(g)\\
2\mbox{ L}
\end{array}$ $\begin{array}{cccc}
\rightarrow & \mbox{CO}_{2}(g) & + & \mbox{H}_{2}\mbox{O}(g)\\
& 8\mbox{ L}
\end{array}$

Jumlah atom C:

$\begin{alignedat}{1}2x & =8\\
x & =4\\
\mbox{C}_{x}\mbox{H}_{2x+2} & =\mbox{C}_{4}\mbox{H}_{(2\times4)+2}\\
& =\mbox{C}_{4}\mbox{H}_{10}.
\end{alignedat}
$

$\begin{array}{ccccc}
\mbox{N}_{2}\mbox{O}_{4} & + & \mbox{NO} & \rightarrow & \mbox{N}_{x}\mbox{O}_{y}\\
25\mbox{ L} & & 50\mbox{ L} & & 50\mbox{ L}\\
1 & & 2 & & 2\\
\mbox{N}_{2}\mbox{O}_{4} & + & \mbox{2NO} & \rightarrow & \mbox{2N}_{x}\mbox{O}_{y}
\end{array}$

Jumlah atom N:

$\begin{alignedat}{1}2+2 & =2x\\
x & =2
\end{alignedat}
$

Jumlah atom O:

$\begin{alignedat}{1}4+2 & =2y\\
y & =3
\end{alignedat}
$

Jadi rumus molekul oksida nitrogen $\mbox{N}_{2}\mbox{O}_{3}$, sehingga

$\begin{aligned}M_{r}\mbox{N}_{2}\mbox{O}_{3} & =(2\times14)+(3\times16)\\
& =76.
\end{aligned}
$

$\begin{aligned}M_{r}\mbox{ rumus empiris CH}_{2} & =A_{r}\mbox{ C}+(2\times A_{r}\mbox{ H})\\
& =12+(2\times1)\\
& =14
\end{aligned}
$

$\begin{alignedat}{1}x & =\frac{M_{r}\mbox{ rumus molekul}}{M_{r}\mbox{ rumus empiris}}\end{alignedat}
$, nilai $x$ haruslah suatu bilangan bulat

$\begin{alignedat}{1}x & =\frac{82}{14}\\
& =5,86
\end{alignedat}
$

$\begin{alignedat}{1}x & =\frac{58}{14}\\
& =4,14
\end{alignedat}
$

$\begin{alignedat}{1}x & =\frac{68}{14}\\
& =4,87
\end{alignedat}
$

$\begin{alignedat}{1}x & =\frac{42}{14}\\
& =3
\end{alignedat}
$

$\begin{alignedat}{1}x & =\frac{54}{14}\\
& =3,86
\end{alignedat}
$

Massa X dalam 20 gram senyawa $\mbox{X}_{2}\mbox{O}_{3}$ adalah:

$\begin{alignedat}{1}\mbox{massa X} & =\frac{36,84}{100}\times20\mbox{ gram}\\
& =7,368\mbox{ gram}\\
\mbox{massa O} & =20\mbox{ gram}-7,368\mbox{ gram}\\
& =12,632\mbox{ gram}
\end{alignedat}
$

$\begin{array}{ccccccc}
\mbox{mol X} & : & \mbox{mol O} & = & 2 & : & 3\\
\frac{\mbox{massa X}}{A_{r}\mbox{ X}} & : & \frac{\mbox{massa O}}{A_{r}\mbox{ O}} & = & 2 & : & 3\\
\frac{7,368}{A_{r}\mbox{ X}} & : & \frac{12,632}{16} & = & 2 & : & 3\\
& & \frac{7,368}{A_{r}\mbox{ X}}\times3 & = & \frac{12,632}{16} & \times & 2\\
& & \frac{22,104}{A_{r}\mbox{ X}} & = & \frac{12,632}{8}\\
& & A_{r}\mbox{ X} & = & 14.
\end{array}$

Kalsium hidroksida akan bereaksi dengan karbon dioksida hasil reaksi, sehingga berkurangnya volume (setelah dimasukkan ke larutan $\mbox{Ca(OH)}_{2}$) merupakan volume $\mbox{CO}_{2}$.

Volume $\mbox{CO}_{2}$= 60 mL

$\begin{aligned}V_{\mbox{total}} & =V_{\mbox{O}_{2}\mbox{Sisa}}+V_{\mbox{CO}_{2}}\\
90\mbox{ mL} & =V_{\mbox{O}_{2}\mbox{Sisa}}+60\mbox{ mL}\\
V_{\mbox{O}_{2}\mbox{Sisa}} & =90-60\mbox{ mL}\\
& =30\mbox{ mL}
\end{aligned}
$

$\begin{aligned}V_{\mbox{O}_{2}\mbox{Reaksi}} & =V_{\mbox{O}_{2}\mbox{Total}}-V_{\mbox{O}_{2}\mbox{Sisa}}\\
& =120\mbox{ mL}-30\mbox{ mL}\\
& =90\mbox{ mL}
\end{aligned}
$

$\begin{array}{ccc}
\mbox{C}_{x}\mbox{H}_{y} & + & \mbox{O}_{2}\\
15\mbox{ mL} & & 90\mbox{ mL}\\
1 & : & 6
\end{array}$$\begin{array}{cccc}
\rightarrow & \mbox{CO}_{2} & + & \mbox{H}_{2}\mbox{O}\\
& 60\mbox{ mL}\\
: & 4
\end{array}$

Persamaan di atas disetarakan, ingat perbandingan volume = perbandingan koefisien (pada tekanan dan suhu yang sama)

$\mbox{C}_{x}\mbox{H}_{y}+6\mbox{O}_{2}$ $\rightarrow4\mbox{CO}_{2}+\mbox{4H}_{2}\mbox{O}$

$\begin{array}{ccccc}
\mbox{jumlah C} & \Rightarrow & x & = & 4\\
\mbox{jumlah H} & \Rightarrow & y & = & 8
\end{array}$

Rumus hidrokarbon $\mbox{C}_{x}\mbox{H}_{y}=\mbox{C}_{4}\mbox{H}_{8}$,

Massa 1 mol = $M_{r}$, Sehingga

$\begin{aligned}M_{r}\mbox{ C}_{4}\mbox{H}_{8} & =(4\times A_{r}\mbox{ C})+(8\times A_{r}\mbox{ H})\\
& =(4\times12)+(8\times1)\\
& =48+8\\
& =56.
\end{aligned}
$

Dalam 0,267 gram $\mbox{C}_{a}\mbox{H}_{b}\mbox{N}_{c}\mbox{O}_{d}$ terdapat 0,3128 gram $\mbox{CO}_{2}$ dan 0,06416 gram $\mbox{H}_{2}\mbox{O}$.

$\begin{aligned}\mbox{Massa C}\mbox{dalam CO}_{2} & =\frac{12}{44}\times0,3128\mbox{ g}\\
& =0,0853\mbox{ g}
\end{aligned}
$

$\begin{aligned}\mbox{Massa H dalam H}_{2}\mbox{O} & =\frac{2\times1}{18}\times0,06416\mbox{ g}\\
& =7,13\times10^{-3}\mbox{g}
\end{aligned}
$

Dalam 0,135 gram $\mbox{C}_{a}\mbox{H}_{b}\mbox{N}_{c}\mbox{O}_{d}$ terdapat 0,0306 gram $\mbox{NH}_{3}$

$\begin{aligned}\mbox{Massa N dalam NH}_{3} & =\frac{14}{17}\times0,0306\mbox{ g}\\
& =0,0252\mbox{ g}
\end{aligned}
$

$\begin{aligned}\mbox{\% N} & =\frac{0,0252}{0,135}\times100\%\\
& =18,67\%
\end{aligned}
$

$\begin{aligned}\mbox{Massa N dalam 0,267g} & =18,67\%\times0,267\mbox{ g}\\
& =0,04984\mbox{ g}
\end{aligned}
$

$\begin{aligned}\mbox{Massa O} & =0,25577-0,0852-0,00713-0,04984\mbox{ g}\\
& =0,1136\mbox{ g}
\end{aligned}
$

$\begin{array}{c}
\mbox{C}\\
\frac{0,0853}{12}\\
7,1083\times10^{-3}\\
1\\
2
\end{array}$ $\begin{array}{cc}
: & \mbox{H}\\
: & \frac{0,00713}{1}\\
: & 7,13\times10^{-3}\\
: & 1\\
: & 2
\end{array}$ $\begin{array}{cc}
: & \mbox{N}\\
: & \frac{0,04984}{14}\\
: & 3,56\times10^{-3}\\
: & 0,5\\
: & 1
\end{array}$ $\begin{array}{cc}
: & \mbox{O}\\
: & \frac{0,1136}{16}\\
: & 7,1\times10^{-3}\\
: & 1\\
: & 2.
\end{array}$

Maka, rumus empirisnya $\mbox{C}_{2}\mbox{H}_{2}\mbox{NO}_{2}.$

$\begin{alignedat}{1}\rho_{\mbox{udara}} & =1,18\mbox{ kg/m}^{3}\\
& =0,00118\mbox{ g/cm}^{3}\\
V_{\mbox{udara(dalam tabung)}} & =200\mbox{ cm}^{3}
\end{alignedat}
$

$\begin{alignedat}{1}m_{\mbox{udara}} & =\rho_{\mbox{udara}}\times V_{\mbox{udara}}\\
& =0,00118\mbox{ g/cm}^{3}\times200\mbox{ cm}^{3}\\
& =0,236\mbox{ gram}\\
m_{\mbox{tabung}} & =m_{\mbox{tabung+udara}}-m_{\mbox{udara}}\\
& =46,051\mbox{ gram}-0,236\mbox{ gram}\\
& =45,815\mbox{ gram}\\
m_{\mbox{C}_{x}\mbox{H}_{y}} & =m_{\mbox{tabung+C}_{x}\mbox{H}_{y}}\mbox{-}m_{\mbox{tabung}}\\
& =46,1739\mbox{ gram}-45,815\mbox{ gram}\\
& =0,3589\mbox{ gram}\\
PV & =nRT\\
PV & =\frac{m}{M_{r}}RT\\
M_{r} & =\frac{mRT}{PV}\\
& =\frac{0,3589\times0,082\times(26+273)}{1\times0,2}\\
& =43,998\\
& =44
\end{alignedat}
$

$\begin{array}{ccccccc}
\mbox{C}_{x}\mbox{H}_{y} & + & \mbox{O}_{2} & \rightarrow & \mbox{CO}_{2} & + & \mbox{H}_{2}\mbox{O}\\
& & & & 39,6\mbox{ gram} & & 13,2\mbox{ gram}
\end{array}$

$\begin{alignedat}{1}\mbox{massa C }\left(\mbox{C}_{x}\mbox{H}_{y}\right) & =\mbox{massa C }\left(\mbox{CO}_{2}\right)\\
\mbox{massa C} & =\frac{\mbox{jumlah atom C}\times A_{r}\mbox{C}}{M_{r}\mbox{ CO}_{2}}\times\mbox{massa CO}_{2}\\
& =\frac{1\times12}{44}\times39,6\mbox{ gram}\\
& =10,8\mbox{ gram}\\
\mbox{massa H }\left(\mbox{C}_{x}\mbox{H}_{y}\right) & =\mbox{massa H }\left(\mbox{H}_{2}\mbox{O}\right)\\
& =\frac{\mbox{jumlah atom H}\times A_{r}\mbox{H}}{M_{r}\mbox{ H}_{2}\mbox{O}}\times\mbox{massa H}_{2}\mbox{O}\\
& =\frac{2\times1}{18}\times13,2\mbox{ gram}\\
& =2,4\mbox{ gram}
\end{alignedat}
$

$\begin{array}{ccc}
\mbox{mol C} & : & \mbox{mol H}\\
\frac{10,8}{12} & : & \frac{2,4}{1}\\
0,9 & : & 2,4\\
3 & : & 8\\
\mbox{RE} & = & \mbox{C}_{3}\mbox{H}_{8}\\
x & = & \frac{M_{r}\mbox{RM}}{M_{r}\mbox{ RE}}\\
& = & \frac{44}{44}\\
& = & 1\\
\mbox{RM} & = & \mbox{C}_{3}\mbox{H}_{8}.
\end{array}$

$\begin{aligned}PV & =nRT\\
1\times36 & =\frac{61,5}{M_{r}}\times0,082\times(27+273)\\
M_{r} & =42
\end{aligned}
$

$\begin{aligned}x & =\frac{M_{r}\mbox{ RM}}{M_{r}\mbox{ RE}}\\
& =\frac{42}{14}\\
& =3
\end{aligned}
$

$\mbox{RM}=\mbox{C}_{3}\mbox{H}_{6}$

Jumlah molekul $\mbox{C}_{3}\mbox{H}_{6}$ dengan massa 2,1 gram:

$\begin{aligned}n & =\frac{2,1}{42}\mbox{ mol}\\
& =0,05\mbox{ mol}
\end{aligned}
$

$\begin{aligned}\mbox{Jumlah molekul} & =n\times6,02\times10^{23}\\
& =0,05\times6,02\times10^{23}\\
& =3,01\times10^{22}
\end{aligned}
$

$\begin{aligned}\mbox{Jumlah atom H} & =6\times3,01\times10^{22}\\
& =1,806\times10^{23}.
\end{aligned}
$