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QII. a)Calculate the percentage composition of the following:-
(H=1, S= 32,O= 16, C= 12) [2]
1.Oxygen in sulphuric acid
2.Carbon in Sugar

b).Derive the empirical formula of the following:- [3]
1.Benzene 2. Sugar 3. Hydrogen peroxide
QI. Calculate the molecular mass of the following:- [5]
3.Sodium zincate [Na=23, Zn=65, O=16]
4.Aluminium nitride [Al=27, N=14]
5.Ferrous chloride [Fe=56 , Cl=35.5]

WITH SOLUTION PLEASE THANK YOU

Sagot :

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Answer:

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a) Percentage Composition:

1. Oxygen in Sulphuric Acid (H2SO4):

- Molecular mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g/mol

- Oxygen's contribution = (4 * 16) / 98 * 100% = 64/98 * 100% ≈ 65.31%

2. Carbon in Sugar:

- There are different types of sugar, but for simplicity, let's consider sucrose (table sugar) with the formula C12H22O11.

- Molecular mass of C12H22O11 = 12(12) + 22(1) + 11(16) = 342 g/mol

- Carbon's contribution = (12 * 12) / 342 * 100% ≈ 42.11%

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b) Empirical Formulas:

1. Benzene (C6H6):

- The empirical formula is the simplest whole number ratio of atoms in a compound, so for benzene, C6H6, the ratio is 1:1, making the empirical formula CH.

2. Sugar (assuming sucrose, C12H22O11):

- The empirical formula of sugar is CH2O.

3. Hydrogen Peroxide (H2O2):

- The empirical formula of hydrogen peroxide is HO.

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QI. Molecular Mass Calculation:

3. Sodium Zincate (Na2ZnO2):

- Molecular mass = 2(23) + 65 + 2(16) = 23 + 23 + 65 + 32 = 143 g/mol

4. Aluminium Nitride (AlN):

- Molecular mass = 27 + 14 = 41 g/mol

5. Ferrous Chloride (FeCl2):

- Molecular mass = 56 + 2(35.5) = 56 + 71 = 127 g/mol

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I hope this breakdown helps.

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