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deff fn [24]
2 years ago
14

Calculate the percentage of sodium and carbon and oxygen in sodium carbonate​

Chemistry
1 answer:
Fantom [35]2 years ago
8 0
Percentage of Sodium [Na] in Sodium Carbonate [Na2CO3] = 46/106 x 100 = 43.40%. Percentage of Carbon [C] in Sodium Carbonate [Na2CO3] = 12/106 x 100 = 11.32%. Percentage of Oxygen [O] in Sodium Carbonate [Na2CO3] = 48/106 x 100 = 45.28% .....
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Answer:

Empirical formula: BH3

Molecular Formula: B2H6

Explanation:

To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:

100% _____ 27 g

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100% ______27 g

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100% _____ 28g

21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen

So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.

The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.

The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.

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