1. determine the percentage composition of al2(co3)3.\n2. a compound consists of 20.32 g of carbon, 5.12 g…

1. determine the percentage composition of al2(co3)3.\n2. a compound consists of 20.32 g of carbon, 5.12 g of hydrogen, and 7.9 g of nitrogen.\n(a) what is the empirical formula?\n(b) what is the molecular formula if the molar mass of the compound is 236.448 g/mol?\nempirical formula:\n|element|c|h|n|\n|--|--|--|--| \n|mass (g)|20.32|5.12|7.9|\n|atomic mass|12.013|1.0083|14.013|\n|mole = mass / atomic mass|20.32 / 12.013 ≈ 1.69|5.12 / 1.0083 ≈ 5.08|7.9 / 14.013 ≈ 0.563|\n|mole ratio = mole / smallest mole|1.69 / 0.563 ≈ 3|5.08 / 0.563 ≈ 9|0.563 / 0.563 = 1|

1. determine the percentage composition of al2(co3)3.\n2. a compound consists of 20.32 g of carbon, 5.12 g of hydrogen, and 7.9 g of nitrogen.\n(a) what is the empirical formula?\n(b) what is the molecular formula if the molar mass of the compound is 236.448 g/mol?\nempirical formula:\n|element|c|h|n|\n|--|--|--|--| \n|mass (g)|20.32|5.12|7.9|\n|atomic mass|12.013|1.0083|14.013|\n|mole = mass / atomic mass|20.32 / 12.013 ≈ 1.69|5.12 / 1.0083 ≈ 5.08|7.9 / 14.013 ≈ 0.563|\n|mole ratio = mole / smallest mole|1.69 / 0.563 ≈ 3|5.08 / 0.563 ≈ 9|0.563 / 0.563 = 1|

Answer

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:

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