Question

The following question was posed on an exam:

An unknown non-metal element (Q) forms two gaseous fluorides of unknown molecular formula. A 3.2g sample of Q reacts with fluorine to form 10.8 g of the unknown fluoride A. A 6.4 g sample of Q reacts with fluorine to form 29.2 g of unknown fluoride B. Using these data only, demonstrate by calculation and explanation that these unknown compounds obey the Low of Multiple Proportions.
A student responded with the following answer: The Law of Multiple Proportions states that when two elements form two or more compounds, the ratios of the masses of the elements between the two compounds are in o simple whole number ratio. So, looking of the data above, we see that the ratio of the moss of element Q in compound A to the mass of element Q in compound B is, which is a simple whole number ratio. This demonstrates that these compounds obey the low of Multiple Proportions. Assess the accuracy of the students answer. In your assessment, you must determine what information is correct or Incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provide logical reasoning where needed.

Answer 1

The answer given by the student is not totally correct.

Step-by-step explanation:

Law of multiple proportions states that an element Q will react with different volume of Fluorine to produce two non-similar compounds. Hence, the ratio of the masses(Fluorine only) needs to be an absolute reduced value.

Given that:

3.2 g of a sample of Q reacts with Fluorine to form 10.8 g of the unknown fluoride A.

This means the mass of Fluorine present in the compound = 10.8 g - 3.2 g = 7.6 g

Thus, 3.2 g of a sample of Q reacts with 7.6 g of Fluorine.

Also, 6.4 g sample of Q reacts with Fluorine to form 29.2 g of unknown fluoride B.

If we divide the samples by (2), we have 3.2 g sample of Q reacting with Fluorine to form 14.6 g of unknown fluoride B.

This means the mass of Fluorine present in the compound = 14.6 g - 3.2 g = 11.4 g

Thus, 3.2 g Q reacted here with 11.4 g fluorine.

Hence, 11.4/7.6 = 1.5 = 3/2

This above therefore satisfies the law of multiple proportions.

This is not aligned with what the student did, what the student did was to relate the amount of Q used to make A and B. Suppose, we start with twice amount of Q, the ratio would have been smaller(i.e. the ratio of Q).

So, this doesn't relate to the law of multiple proportions.

The law of multiple proportions is specifically concerned with the mass of the element rather than Q that can react with Q. Therefore, there is more reason to relate the two samples of equal masses of Q that react with different masses of Fluorine. The ratios of Fluorine will then be small whole numbers.