Question

1. determine the percentage composition of al2(co3)3.
2. a compound consists of 20.32 g of carbon, 5.12 g of hydrogen, and 7.9 g of nitrogen.

(a) what is the empirical formula?
(b) what is the molecular formula if the molar mass of the compound is 236.448 g/mol?
empirical formula:

element	c	h	n
atomic mass	12.013	1.0083	14.013
mole = mass / atomic mass	20.32 / 12.013 ≈ 1.69	5.12 / 1.0083 ≈ 5.08	7.9 / 14.013 ≈ 0.563
mole ratio = mole / smallest mole	1.69 / 0.563 ≈ 3	5.08 / 0.563 ≈ 9	0.563 / 0.563 = 1

Explanation:

Step1: Calculate moles of each element

Moles of carbon ($n_{C}$) = $\frac{20.32\ g}{12.01\ g/mol}\approx1.692\ mol$
Moles of hydrogen ($n_{H}$) = $\frac{5.12\ g}{1.008\ g/mol}\approx5.08\ mol$
Moles of nitrogen ($n_{N}$) = $\frac{7.9\ g}{14.01\ g/mol}\approx0.564\ mol$

Step2: Determine mole - ratio

Divide each mole value by the smallest mole value (0.564 mol).
Mole - ratio of carbon = $\frac{1.692\ mol}{0.564\ mol}=3$
Mole - ratio of hydrogen = $\frac{5.08\ mol}{0.564\ mol}\approx9$
Mole - ratio of nitrogen = $\frac{0.564\ mol}{0.564\ mol}=1$
Empirical formula is $C_{3}H_{9}N$.

Step3: Calculate empirical - formula mass

Empirical - formula mass ($M_{EF}$) = $3\times12.01\ g/mol+9\times1.008\ g/mol + 14.01\ g/mol$
$M_{EF}=36.03\ g/mol+9.072\ g/mol+14.01\ g/mol = 59.112\ g/mol$

Step4: Calculate the multiple (n)

$n=\frac{\text{Molar mass}}{\text{Empirical - formula mass}}=\frac{236.448\ g/mol}{59.112\ g/mol}=4$

Step5: Determine molecular formula

Molecular formula = $(C_{3}H_{9}N)_4 = C_{12}H_{36}N_{4}$

Answer:

$C_{12}H_{36}N_{4}$