Two elements A (atomic mass = 75) and B (atomic mass = 16) combine to ...
Molecular weight of A2B3 = 75x2+16x3=198
198g of A2B3 = 150g of A
∴ 10g of A2B3 = (150x100)/198 = 75.08g of A
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Two elements A (atomic mass = 75) and B (atomic mass = 16) combine to ...
Given Information:
- Atomic mass of element A = 75
- Atomic mass of element B = 16
- Percentage by weight of A in the compound = 75.08%
Explanation:
To determine the formula of the compound, we need to consider the concept of molar mass and the law of definite proportions.
1. Molar Mass:
The molar mass of an element or compound is the mass of one mole of that substance. It is calculated by summing the atomic masses of all the atoms present in the formula.
2. Law of Definite Proportions:
The law of definite proportions states that a chemical compound always contains the same elements in the same proportion by mass.
Steps to Determine the Formula:
1. Assume we have 100g of the compound.
2. Given that the percentage by weight of A in the compound is 75.08%, it means that 75.08g of the compound is made up of element A.
3. The remaining weight of the compound is 100g - 75.08g = 24.92g, which is made up of element B.
4. Now, we need to determine the number of moles of each element in the compound.
- Moles of A = mass of A / molar mass of A = 75.08g / 75g/mol = 1 mole
- Moles of B = mass of B / molar mass of B = 24.92g / 16g/mol = 1.56 moles
5. Since the compound follows the law of definite proportions, the ratio of moles of A to B must be a whole number.
6. Dividing the moles of B by the moles of A, we get 1.56 / 1 = 1.56. Since this is not a whole number, we need to multiply both moles of A and B by the smallest whole number to obtain a whole number ratio.
7. Multiplying both moles of A and B by 2, we get 2 moles of A and 3.12 moles of B.
8. Rounding off the moles to the nearest whole number, we have 2 moles of A and 3 moles of B.
9. Therefore, the formula of the compound is A2B3, which is option (b).
Hence, the correct answer is option (b) A2B3.