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Three samples of a solid substance composed of elements A and Z were prepared. The first contained 4.31 g A and 7.70 g Z. The second sample was 35.9% A and 64.1% Z. It was observed that 0.718 g A reacted with Z to form 2.00 g of the third sample. Show that these data illustrate the law of definite composition.

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Answer:

The percentage composition of the elements of the compound in the three samples is the same.

Step-by-step explanation:

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.

Sample A:

Mass of A = 4.31 g; mass of Z = 7.70 g

Total mass of sample = 12.01

Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %

Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %

Sample B:

Percentage mass of A in the sample = 35.9 %

Percentage mass of Z in the sample = 64.1 %

Sample C:

Mass of A = 0.718 g; Total mass of sample = 2.00 g

mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g

Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %

Percentage mass of Z in the sample = (1.282 * 100)/2.00 = 64.1 %

From the calculations, it can be seen that the percentage composition of the elements in the compound is the same for the three samples.

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