2009.10.12 SAMPLE PROBLEM

2009.10.12 SAMPLE PROBLEM - This means we will get 11 moles...

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Sample Problem from 10/12/09 Lecture Problem E.9, part b) How many O atoms are in 5.15 g of MgSO 4 . 7H 2 O? Step 1 : Determine molar mass of MgSO 4 . 7H 2 O. Molar mass of MgSO 4 . 7H 2 O = molar mass of Mg + molar mass of sulfur + 4 (molar mass of O) + 7 (molar mass of water) = 246.51 g/mol Step 2 : How many moles of MgSO 4 . 7H 2 O are there in 5.15 g? Moles of MgSO 4 . 7H 2 O = 5.15 g 1 mol 246.51 g " # $ % = 0.0209 mol Step 3: In each mole of MgSO 4 . 7H 2 O, how many moles of O atoms are there? In other words, if we break up 1 mole of MgSO 4 . 7H 2 O, how many moles of O will we have? There are 7 O atoms in the 7 water molecules and 4 O atoms in the sulfate anion.
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Unformatted text preview: This means we will get 11 moles of O atoms for every mole of MgSO 4 . 7H 2 O. Mathematically this is 11 mol O/ 1 mol MgSO 4 . 7H 2 O. Step 4: How many moles of O atoms are there in 5.15 g (0.0209 mol) of MgSO 4 . 7H 2 O? Moles of O atoms = 0.0209 mol MgSO 4 . 7H 2 O (11 mol O / 1 mol MgSO 4 . 7H 2 O) = 0.230 mol O atoms Step 5: How many O atoms? Using our conversion factor, Avogadros constant, we find that we have # of O atoms = 0.230 mol 6.022 x 10 23 1 mol " # $ % & = 1.38 x 10 23 O atoms...
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This note was uploaded on 04/29/2010 for the course CHEM CHEM 6A taught by Professor Czarkowski during the Spring '07 term at UCSD.

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