**Question 1:** Find out the value of molecular weight of the given compounds:

(i) CH_{4} (ii)H_{2}O (iii)CO_{2}

**Answer:**

(i)CH_{4} :
Molecular weight of methane, CH_{4}

= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)

= [1(12.011 u) +4 (1.008u)]

= 12.011u + 4.032 u

= 16.043 u

(ii) H_{2}O :

Molecular weight of water, H_{2}O

= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u +16.00 u

= 18.016u

So approximately

= 18.02 u

(iii) CO_{2} :

= Molecular weight of carbon dioxide, CO_{2}

= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)

= [1(12.011 u) + 2(16.00 u)]

= 12.011 u +32.00 u

= 44.011 /u /

So approximately

= 44.01u

Question 2: Sodium Sulphate (Na_{2}SO_{4}) has various elements, find out the mass percentage of each element.

**Answer:**

Now for Na_{2}SO_{4}.
Molar mass of Na_{2}SO_{4}

= [(2 x 23.0) + (32.066) + 4(16.00)]
=142.066 g

Formula to calculate mass percent of an element =

Mass of that element in the compound Molar mass of the compound × 100

Therefore, Mass percent of the sodium element:

= 46.0g142.066g×100
= 32.379
=32.4%

Mass percent of the sulphur element:

=32.066g142.066g×100
= 22.57
=22.6%

Mass percent of the oxygen element:

= 64.0g142.066g×100
=45.049
=45.05%

**Question 3:**Find out the empirical formula of an oxide of iron having 69.9% Fe and 30.1% O2 by mass.

**Answer:**

Percent of Fe by mass = 69.9 % [As given above]

Percent of O2 by mass = 30.1 % [As given above]

Relative moles of Fe in iron oxide:

= percent of iron by mass Atomic mass of iron

= 69.955.85
= 1.25

Relative moles of O in iron oxide:

= percent of oxygen by mass Atomic mass of oxygen

= 30.116.00
= 1.88

Simplest molar ratio of Fe to O:

= 1.25: 1.88

= 1: 1.5
≈ 2: 3

Therefore, empirical formula of iron oxide is Fe_{2}O_{3}.

**Question 4:**Find out the amount of CO_{2} that can be produced when

(i) 1 mole carbon is burnt in air.

(ii) 1 mole carbon is burnt in 16 g of O_{2}.

(iii) 2 moles carbon are burnt in 16 g O_{2}.

**Answer:**

(i) 1 mole of carbon is burnt in air.

C+O_{2}→CO_{2}

1 mole of carbon reacts with 1 mole of O2 to form one mole of CO_{2}.

Amount of CO_{2} produced = 44 g

(ii) 1 mole of carbon is burnt in 16 g of O_{2}.

1 mole of carbon burnt in 32 grams of O_{2} it forms 44 grams of CO_{2}.

Therefore, 16 grams of O_{2} will form 44×1632

= 22 grams of CO_{2}

(iii) 2 moles of carbon are burnt in 16 g of O_{2}.

If 1 mole of carbon are burnt in 16grams of O_{2} it forms 22 grams of CO_{2}

Therefore, if 2 moles of carbon are burnt it will form

= 2×221

= 44g of CO_{2}

**Question 5:**Find out the mass of CH_{3}COONa(sodium acetate) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of CH3COONa is 82.0245gmol^{−1}

**Answer:**

0.375 Maqueous solution of CH_{3}COONa

= 1000 mL of solution containing 0.375 moles of CH_{3}COONa

Therefore, no. of moles of CH_{3}COONa in 500 mL

= 0.3751000×1000

= 0.1875 mole

Molar mass of sodium acetate = 82.0245gmol^{−1}

Therefore, mass that is required of CH_{3}COONa

= (82.0245gmol^{−1})(0.1875mole)

= 15.38 gram

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