5 Determine the number of moles of each compound a 123g CuCl

5. Determine the number of moles of each compound: a. 12.3g CuCl b. 0.45g CHal (I is iodine) c. 0.357mg NH2CH COOH d. 1.4kg CaHPO4 6. Determine the empirical formula of each compound from percent composition: a. b. c, Cr 19.64% and Cl 80.36% Cu 39.81%, S 20.09% and O 40.10% C 32.00%, O 42.63%, N 18.66% and H 6.71%

Solution

5.A) Molecular weight of CuCl₂=134 gm

so, 134g CuCl_{2 = 1 mole}

_{12.3 gm} CuCl_{2 = 0.091 mole}

B)Molecular weight of CH₃I=134 gm

SO, 134g CH₃I = 1 mole

0.45 g CH₃I = 0.00335 mole

c) Molecular weight of NH₂CH₂COOH =75GM

SO, 75 gm NH₂CH₂COOH = 1 mole

0.357 mg=0.357 * 10 ^-3 gm NH₂CH₂COOH = 0.00476 * 10^-3 mole

D) Molecular weight of CaHPO4 = 136 gm

so, 136gm NH₂CH₂COOH = 1 mole

1.4kg = 1400gm NH₂CH₂COOH= 10.29 mole

6) a) Assume 100 g of the compound is present. This changes the percents to grams:

Cr 19.64 g
Cl 80.36 g

Convert the masses to moles:

Cr 19.64 g / 52 g/mol = 0.26 mol
Cl 80.36 g / 35.5 g/mol = 2.26 mol

Divide by the lowest, seeking the smallest whole-number ratio:

Cr 0.26 /0.26 = 1
Cl 2.26 /0.26 = 9

empirical formula: CrCl₉

b) By same procedure empirical formula : CuSO₄

C) By same procedure empirical formula : C₂ONH₃