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Fall 2008 |
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Discussion Modules |
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1. How many moles of chlorine are there in 1.5 moles of iron(III) chloride?
First we have to determine the formula of iron(III) chloride. Iron(III) = Fe3+ while chloride is an elemental ion in Group VII:
IA IIA IIIA IVA VA VIA VIIA VIIIA IIIB IVB VB VI VIIB VIIIB IB IIB Cl Therefore it has a charge of 7-8 = -1 so chloride = Cl-.
Since the charge on any compound MUST be zero there are three chlolrides/iron(III), giving us a formula of FeCl3.
From the formula we can see that there are three chlorines for each formula, therefore there are three moles of chlorine for each mole of iron(III) chloride, giving:
(1.5 mol FeCl3)(3 mol Cl/mol FeCl3) = 4.5 mol Cl
Note that dimensional analysis will give this expression, and the significant figures (2) are determined by the measured number of 1.5, NOT the theoretical ratio.
2. How many moles of hydrogen are there in 0.763 moles of ammonium sulfide?
First we have to determine the formula of ammonium sulfide. Our vast knowledge of chemical names gives us the formula of ammonium ion as NH4+ while sulfide is an elemental ion in Group VI
IA IIA IIIA IVA VA VIA VIIA VIIIA IIIB IVB VB VI VIIB VIIIB IB IIB S Therefore it has a charge of 6-8 = -2, so sulfide = S2-.
Since the charge on any compound MUST be zero there are two ammonium ions/sulfide, giving us a formula of (NH4)2S. .
From the formula we can see that there are 2x4 = 8 hydrogens for each formula, therefore there are eight moles of hydrogen for each mole of ammonium sulfide, giving:
(0.763 mol (NH4)2S)(8 mol H/(NH4)2S) = 6.104 = 6.10 mol H
Note that dimensional analysis will give this expression, and the significant figures (3) are determined by the measured number of 0.763, NOT the theoretical ratio.
3. How many moles of aluminum are there in 2.34 g of aluminum foil?
The first thing is to look up the atomic weight of aluminum on the Periodic chart:
IA IIA IIIA IVA VA VIA VIIA VIIIA IIIB IVB VB VI VIIB VIIIB IB IIB 13
Al
26.98
And we see that aluminum weighs 26.98g/mol. Using dimensional analysis we need moles on top in the answer. Multiplying by the mass will then cancel the grams, giving:
{(1mol Al)/(26.98 g Al)}x{2.34 g Al} = 8.673x10-2 = 8.67x10-2 mol
4. How many moles of oxygen are there in a 76.94 g sample of air containing 19% oxygen?
First we need to find the mass of the oxygen = (76.94g)(0.19) = 14.62g O2
The atomic mass of oxygen from the Periodic table = 16.00g O/mol O. Using dimensional analysis as in the previous problem,
(1 mol O/16.00g O)(14.62g O) = 0.9137 = 0.91 mol
Note the significant figures are controlled by the initial percentage with 2 s.f.
5. How many grams of silver are there in 10.265 mol of silver phosphide?
Since the formula of a compound gives the ratio of its atoms, and therefore the ratio of moles, we need to first find the formula of the compound.
Looking at the Periodic table we see that silver has a charge of +1 and the elemental ion of phosphorus has a charge of 5-8 = -3. The silver phosphide thus has the formula Ag3P and there are three moles of silver for each mole of silver phosphide:
(3 mol Ag/mol Ag3P)(10.265 mol Ag3P) = 30.795 mol Ag
Multiplying the moles by the atomic weight of silver from the Periodic table,
(30.795 mol Ag)(107.9g Ag/mol Ag) = 3.3228x103 = 3.323x103 = 3.323x103 g silver
6. How many grams of hydrogen would be needed to make 1.023 mol of ethanol (CH3CH2OH)?
First we need to find the ratio of moles hydrogen needed to make a mole of ethanol. We can either write a balanced equation, or simply count the number of hydrogens in the formula, we will count.
Looking at the formula, the number of hydrogens = 3 + 2 + 1 = 6
Next we need to find th eatomic weight of hydrogen from the Periodic table = 1.008 g/mol. Using dimensional analysis and plugging in our numbers we find,
(1.008 g H/mol H)(6 mol H/mol ethanol)(1.023 mol ethanol) = 6.1871 = 6.187 g hydrogen
© R A Paselk
Last modified 23 October 2008