You take three compounds, each consisting of two elements...

You take three compounds, each consisting of two elements from the set (X, Y, and Z) and decompose them to their respective elements. To determine the relative masses of X, Y, and Z, you collect and weigh the elements, obtaining the following data: Elements in Compound Masses of Elements 1. X and Y X=0.4g, Y = 4.2 g 2. Y and Z Y=1.4g, Z = 1.0 g 3. X and Y X=2.0 g, Y = 7.0 g Assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ. (a) What are the relative masses of X, Y, and Z? (Enter your answers as the smallest whole numbers.) Relative mass of X = Relative mass of Y = Relative mass of Z = (b) What is the chemical formula of the compound in experiment 3? Formula: (c) If you decompose 22.0 g of compound XY, how much of each element is present? (Enter your answers to one decimal place.) Mass of Y = g Mass of X = g

Transcript

00:08 If we take three compounds consisting of two elements, x, y, and or z, and decompose them to their respective elements, so we have the data in the table given.

00:19 For part a, what are the assumptions needed to solve this problem? to solve this problem, we need to assume that the compounds obey the law of multiple.

00:55 Proportions.

01:01 For part b, one of the relative masses of x, y, and z.

01:11 So we're going to start with compound one.

01:18 And for compound one, we're told that the mass of x is 0 .4 grams.

01:25 And the mass of y is equal to 4 .2 grams.

01:37 And therefore, we can calculate the amounts of x per y, which is equal to 0 .4 grams over 4 .2 grams, which is equal to 1 over 0 .095.

02:16 If we take compound 2, we have the mass of y and the mass of z, 1 .4 grams for y.

02:42 Mass of z is 1 .0 grams.

02:48 And mass of z per y, the amount of z per y would be 1 .0 grams over 1 .4 grams, which would equal 1 over 0 .71.

03:10 And for compound 3, we've got the mass of x and the mass of y.

03:31 Mass of x is 2 .0 grams.

03:34 The mass of y is 7 .0 grams.

03:40 The amounts of x per y is equal to 2 .0 grams over 7 .0 grams, which is equal to 1 over 7 .0 grams, which is equal to 1 over over the point to nine.

04:09 So these would be the relative masses of x, of y, and of z relative to each of the elements.

04:23 So for part c, what are the chemical formulas of each of the three compounds? so using the law of multiple proportions, we can take compound one divided by compound three, which is equal to compound one, that's 0 .095, over compound 3, which is 0 .029.

05:06 Or point, sorry, 0 .29.

05:15 And we find that this is equal to 1 third.

05:19 So we can create a chemical formula here.

05:27 For compound one, we see that the formula would be xy.

05:43 And for compound three, we would get x3y.

05:57 Now using the same idea, we'll bring in compound two.

06:04 So compound one, over compound 2.

06:17 This would be equal to 0 .095 over 0 .71, and this would be equal to 1 over 7 .5.

06:28 Getting rid of the decimal, multiplying by 2 would be 2 .15th...

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