Anonymous

Calculate each of the following quantities:

(a)Mass in grams of 0.57 mol of KMNO4

(b)Moles of O atoms in 8.18 g of Mg(NO3)2

(c)Number of O atoms in 8.1x10^-3 g of CuSO4 * 5H2O

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• Anonymous

(a) 0.57 mol KMNO4 x 158 g KMNO4 = 9.0 x 10 g KMNO4

1 mol KMNO4

1 mol is equivalent to the molar mass of the element so set up a ratio like above. As a rule, your units should cancel to the units you want

(b) 8.18 g Mg(NO3)2 x 1 mole Mg(NO3)2 x 6 mol O =

148 g Mg(NO3)2 1 mol Mg(NO3)2

= 0.332 mol O

These kind of problems are just setting up ratios until you end up with the units you want. You have to convert the Mg(NO3)2 to moles and set up a ratio between these moles and the number of moles of O in Mg(NO3)2

(Multiply the top and multiply the bottom values separately. And then divide the top value by the bottom value to solve)

(c) 8.1x10^-3 g of CuSO4 * 5H2O x 1 mol CuSO4 * 5H2O

250 g CuSO4 * 5H2O

x 9 mol O x (6.022 x 10^23) atoms O = 1.8 x 10^20

1 mol CuSO4 * 5H2O 1 mol O

This problem is very similar to problem (b), after you find out the number of moles of O atoms in CuSO4 * 5H2O, you must then multiply that number of moles by 6.022 x 10^23 atoms (or molecules). This is Avogadro's number.

I hope this helps and check my math.

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1) Remember that on a periodic table, the atomic mass (usually found at the bottom middle of the square) is the mass in grams of ONE MOLE of that element. For example, the mass in grams of one mole of hydrogen is 1.01 grams. In this problem, you don't have an element, but a compound (more than one element chemically connected). For these, just add all of the atomic masses together (39.1 + 54.9 + 16 + 16 + 16 + 16). There are four sixteens because there are four moles of oxygen (the O4). Adding these together gives you the atomic mass for one mole of KMnO4, 158 g. But you only want .57 mol, so multiply 158 g by 0.57 to get your answer, 90.06 g.

2) This is almost the same problem, but in reverse. You have the amount of grams in the compound, which we can convert to moles.

Atomic mass of Mg(NO3)2: 24.3 + 2(14 + 16 + 16 + 16) = 24.3 + 2(62) = 24.3 + 124 = 148.3 g

So, if you have 8.18 g of this compound, and a mole is 148.3 g, then the # of moles you have is 8.18/148.3, or .055 moles.

The number of moles in a compound are the same as the number of moles in each of the elements. Thus, there is .055 moles of Mg, etc. But, there are six times as many moles of O, so there are actually .33 moles of O.

3) I don't understand why you threw in 5H20. Because I don't want to give you the wrong answer, I'm not going to answer at all, unless you can clarify.

Good luck!

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• (a) for that you need the MW of KMnO4 which is 158g/mole

.57mol * 158g/mole = 90g of KMnO4

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• a.KMNO4 molar mass 158.034 g/mol so multiply that by .57 = 90.08 grams

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