Question

Answer the questions completely with complete calculations.

1. How many grams of a sample of NaNO3 should be dissolved in a 150.0 mL of water to give a 0.22 M solution?

2. A certain alcohol contains only three elements C, H & O. Combustion of 50.00 g sample of the alcohol produced 95.50 g of CO2 and 58.70 g of H2O. What is the empirical formula of the alcohol?

Answer 1

1) Let us find first molar mass of NaNO3,

Na, N, O have atomic masses 23, 14, 16 respectively

So. molar mass of NaNO3 = 23 + 14 + (16 x 3) = 85.

According to mole concept, 85 g of NaNO3 = 1 mol NaNO3.

NaNO3 is strong electrolyte hence,

85 g NaNO3 = 1 M NaNO3 ............... in 1000 mL solution

Say 'A' g NaNO3 = 0.22 M NaNO3............. in 1000 mL solution.

so, 'A' = 0.22 / 85 = 2.59 x 10^-3 g NaNO3...........in 1000 mL solution.

so, 1000 mL NaNO3 solution (or water) = 2.59 x 10-3 g NaNO3................(for 0.22 M solution)

then, 150 mL water = say 'B' g NaNO3

implies, 'B' = (150 x 2.59 x 10-3) / 1000

'B'= 3.88 x 10^-4 g NaNO3.

= 0.388 x 10^-3 g

= 0.388 mg.

Hence, 0.388 mg NaNO3 in 150 mL will give 0.22 M NaNO3 solution.

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2) Let the unknown alcohol contains 'p' number of C, 'q' number of H atoms, and 'r 'number of O atoms. Hence the empirical formula of unknown alcohol be C_pH_qO_r.

The balanced chemical equation of combustion of an unknown alcohol is,

C_pH_qO_r + nO2 ---------------> pCO2 + (q/2) H2O. (q/2 for H2O as 2 H per one O atom required)

(12p +1q + 16r) g '32n' g '44p' g 18 x (q/2) g

-//- '44p' g '9q' g

50 g '32n' g 95.50 g 58.70 g

so, we can write,

44p = 95.50 and 9q = 58.70

p = 2.170 q = 6.52

we can also have, 12p + 1q + 16r = 50 (= doesn't means equal just corresponadancy)

Let us put p, q values solve it for r,

12x2.17 + 6.52 + 16r = 50

32.56 + 16r = 50

16r = 50-32.56

16r = 17.44

r = 17.44/16

r = 1.09

We have p = 2.17, q = 6.52, r = 1.09....(these are theoretically atom numbers but atom number cannot be fractional rationalize them)

i.e. p : q : r = 2.17 : 6.52 : 1.09

dividing the ration through out by 1.09 (a smallest of all)

p : q : r = 2.17/1.09 : 6.52/1.09 : 1.09/1.09

p : q : r = 2 : 6 : 1...................(little aproximated again thats fine but)

So unknown alcohol contain 2 C, 6 H, 1 O atom

So, empirical formula is C2H6O.