Stoichiometry and Gas Laws 1
1. What is a Mole?
Stoichiometry problems are based on moles, so let's start with that. For our purposes the number of moles in a sample can be calculated in four different ways.
1. A mole is 6.02´1023 molecules (or atoms or ions) of something. This is Avogadro's Number.
How many moles are equal to 1000 molecules of carbon dioxide?
1000 molecules × 1 mole6.02×1023molecules=1.66×10-21 moles
2. A mole is the molecular weight of something expressed in grams.
How many moles are in 2.56 g of carbon dioxide?
2.56 g × 1 mole44.01 g=0.0582 moles
Sometimes this definition is used for compounds like NaCl or Cu which don't exist as molecules and, therefore, don't have molecular weights. Use the formula weight or the atomic weight instead.
3. The number of moles of a gas is calculated using the ideal gas law PV = nRT. Here "P" is the pressure; "V" is the volume, typically in liters; "n" is the number of moles; "R" is a constant (the gas constant); "T" is the temperature (in Kelvin not Celsius). The value of R, which depends on the units of pressure and volume, will be given to you. However, the most common values are 0.08206 L × atm/mol × K and
8.314 L × kPa/mol × K.
How many moles of carbon dioxide occupy 2.55 L at 25° C and 102.3 kPa?
n= PVRT= 102.3×2.558.314×298=0.105 moles
For the very important case of 0° C and 1 atm pressure (referred to as "standard temperature and pressure" or "STP") one mole of gas occupies 22.4 L. This is referred to as the "molar volume." It is unlikely that you will see a calculation using the molar volume in a problem on the AP test, but much more likely that you will see such a calculation in a multiple choice question.
4. For a solution the number of moles is equal to the molarity times the volume (in liters!).
How many moles of chloride are in 260 mL of 0.120 M CaCl2?
moles = 2 ´ M V = 2 ´ 0.120 M ´ 0.260 L = 0.0624 moles
2. Simple Mole Problems
The empirical formula
This is the simplest formula; it is not necessarily the correct molecular formula. For example the empirical formula for C2H6 is CH3. The easiest way of doing this calculation is to assume that there are 100 g of compound. This makes an element's percentage equal to its weight. Take the weight of each element and divide by the element's atomic weight. (Don't round the answer!) This gives you the moles of element in 100 g of compound. Now take each of these numbers and divide by the smallest number. This will give you the mole ratios, which will give you the coefficients in the formula. If one of the numbers is a fraction, you will have to multiply by something to clear it.
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A compound containing only carbon and hydrogen is found to contain 18.3% hydrogen. What is its empirical formula?
Assume 100 g of compound.
C=81.7 g×1 mole12.01 g=6.80 moles
H=18.3 g×1 mole1.008 g=18.2 moles
HC= 18.26.80=2.67
However CH2.67 is not likely. You need to clear the fraction, which you can do by multiplying by 3. This will give you C3H8.
Percent composition
This is easy to determine. Divide the total atomic weight of each element by the molecular weight of the compound.
What is the elemental percent composition of ethanol, C2H5OH?
C = 2 ´ 12.01 = 24.02 C = 24.02/46.07 = 52.1%
H = 6 ´ 1.008 = 6.048 H = 6.048/46.07 = 13.1%
O = 1 ´ 16.00 = 16.00 O = 16.0/46.07 = 34.7%
______
46.07
The molecular weight of a gas
This can be calculated if you recall that molecular weight is the number of grams in a mole. For that reason it is often called the "molar mass." So
Calculate the molecular weight of a gas for which 1.25 g occupies 320 mL at 95.0 kPa and 20° C.
MW= 1.25 g(95.0×0.320)(8.314×293)=100
The density of a gas
This is the number of grams in a liter. To calculate this you must, again,
recall that molecular weight is the number of grams in a mole.
Density= massvolume=(PVRT)×MWV= P×MWRT
What is the density (in g/L) of HCl gas at 24.0° and 98.5 kPa?
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3. Stoichiometry Problems
Simple Stoichiometry Problems
In most cases you are given information about one compound in a reaction. You convert this to moles, use the balanced equation to convert moles of one compound to moles of another, and then convert moles to whatever you want. This works because the number of molecules in a balanced reaction is proportional to the number of moles of that same material.
How many liters of O2 at 100.2 kPa and 25° C are formed from 2.55 g of KClO3 in the reaction:
2 KClO3® 2 KCl + 3 O2
V= nRTP= 32 × 2.55122.55× 8.31 × 298100.2=0.771 L
Limiting reagent problems
These are like the above; but you are given the quantities of two reactants instead of one. One of the reactants is in excess and the other reactant (the limiting reagent) gets used up. So these problems are done in two steps. First you use the balanced equation to determine which reagent is limiting; then you use the quantity of limiting reagent to determine the quantity of product. The best way to do this is to divide the number of moles of each reactant by its coefficient in the equation. The smaller of the two numbers corresponds to the limiting reagent. Get the number of moles of product by multiplying this number by the coefficient of the product.
Consider the reaction 2A + 3B ® 4C. Calculate the number of moles of A (Divide the grams of A by its molecular weight) and divide by 2 (the coefficient of A). Then calculate the moles of B and divide it by 3. Whichever compound gives the smaller number is limiting. Now take this number and multiply it by 4 (the coefficient of C) to get the moles of C. If you want the mass of C, multiply the moles by the molecular weight.
A piece of aluminum weighing 0.145 g is placed in 750 mL of 0.0200 M HCl. How many mL of dry hydrogen, measured at 25° C and 1.02 atm, are formed?
Al + 3 HCl ® H2 + Al3+ + 3 Cl–
Al = 0.145 g ´ 1 mol/26.98 g = 0.00537 moles
HCl = 0.750 L ´ 0.0200 mol/L = 0.0150 moles
There is one Al atom in the equation; so we divide 0.00537 moles of Al by 1 and get 0.00537. There are 3 moles of HCl in the equation; so we divide 0.0150 by 3 and get 0.00500 moles. The HCl is limiting and the moles of H2 = 3/2 ´ 0.00500 moles = 0.00750 moles.
V= nRTP= 0.00750×0.08206×2981.02=0.180 L=180 mL
4. Partial Pressures (Dalton's Law)
Most partial pressure problems are simple; the total pressure is the sum of the partial pressures. Just remember that the moles of a substance are proportional to its partial pressure. However, there is one particular case which is a bit more difficult. That is where you calculate the amount of gas collected when the gas is wet (collected over water). In such a case you assume that the gas is saturated with water and that the partial pressure of water is equal to its equilibrium vapor pressure at that temperature -- something you can look up in a table.
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A 0.885 g sample of aluminum reacts with acid to form hydrogen. What volume of wet gas will be collected at 25° C and 99.9 kPa? The vapor pressure of water at 25° C is 3.2 kPa. The reaction is:
2 Al + 6 H+® 2 Al3+ + 3 H2
This may be a bit confusing, so we'll do it in two steps.
Moles H2=0.885 g×32×1 mole27.0 g=0.04917 moles
While the pressure of the gas is 99.9 kPa, the pressure of dry hydrogen is only 96.7 kPa.
V= nRTP= 0.04917×8.314×29896.7=1.26 L
5. Simple Mixture Problems
In a simple mixture problem one component of a mixture reacts. An example of this would be a percent purity problem. Typically you use the amount of product to determine the amount of reactant and, thus, the percentage of reactant in the mixture.
Percentage of Reactant
A 4.00 g sample of a mixture of sodium carbonate and sodium chloride is added to excess HCl solution. After drying, the carbon dioxide produced is found to occupy 335 mL at 21° C and 724 torr. Calculate the percentage of sodium carbonate in the mixture. The reaction is:
2 H+ + Na2CO3® 2 Na+ + H2O + CO2
CO2=PVRT=724760×0.3350.08206×294=0.01323 moles
Na2CO3=0.01323 moles CO2×1 mole Na2CO31 mole CO2×105.99 g Na2CO31 mole Na2CO3=1.402 g
%Na2CO3=massNa2CO3 mass sample=1.402 g4.00 g=35.1%
Treating a Compound as a Mixture
In this case the "impurity" is the non-reactive part of the compound.
A 7.500 g sample of a compound containing only carbon and hydrogen is burned to produce 3.794 grams of water. What is the empirical formula of the compound?
Moles of H=3.794 g H2O×1 mole18.02 g×2 moles H1 mole H2O=0.4216 moles
Mass of H=0.4216 moles×1.008 g1 mole=0.4242 g
Mass of C=7.50-0.4242=7.075 g
Moles of C=7.075 g×1 mole12.01 g=0.589 moles
CH=0.5890.4216=1.397=1.4
So the compound is C7H5.
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Percent Purity
An impure sample of potassium metal weighing 3.25 g reacts with water to form 565 mL of hydrogen, measured dry and at 25° C and 98.0 kPa. What is the purity of the potassium. The reaction is:
2 K + 2 H2O ® 2 K+ + H2 + 2 OH–
Moles H2=PVRT=98.0×0.5658.314×298=0.02236
Mass of K=0.02236 moles H2×2 moles K1 mole H2×39.10 g K1 mole K=1.748 g
Purity=mass of Kmass of sample=1.748 g3.25 g=53.8%
6. Mixture problems with two reactive components
These involve a mixture in which both components react to form a product. Usually you set up two simultaneous equations, basing one on the amount of mixture and the other on the amount of product.
A magnesium/aluminum alloy reacts with aqueous HCl to give hydrogen gas. A 0.105 g sample of this alloy reacts to form 116 mL of hydrogen, collected at 20° C and 742 torr. What is the percentage of aluminum in the sample? The reactions are:
Mg + 2 H+® Mg2+ + H2
2 Al + 6 H+® 2 Al3+ + 3 H2
Moles H2=PVRT=742760×0.1160.08206×293=0.00471 moles
In setting up these equations, the variables Al and Mg are both expressed in grams.
Equation 1 -- Moles of Product
H2=32×Al×126.98+Mg×124.31=0.00471
0.05560 ´Al + 0.04114 ´Mg = 0.00471
1.351 ´Al + Mg = 0.1145
Equation 2 -- Mass of Reactants
Al + Mg = 0.105
Subtracting Equation 2 from Equation 1
0.351 ´ Al = 0.0095
Al = 0.0271
= 26%
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ΔP Problems
These are mixture problems in which you use changes in the pressure exerted by a gas mixture to calculate how many moles of each compound are involved in a reaction.
A mixture containing oxygen and a small amount of hydrogen exerts a pressure of 1.02 atm at 18° C. The hydrogen reacts completely, forming water vapor according to the reaction:
2 H2(g) + O2(g) ® 2 H2O(g)
If the final gas mixture exerts a pressure of 0.98 atm, again at 18° C, what is the pressure of hydrogen in the original mixture?
The key to this problem is that the pressure of the system changes by 0.04 atm. Why does it change? Look at the reaction stoichiometry. Two moles of hydrogen react with one mole of oxygen to form 2 moles of water vapor. Thus for every two moles of hydrogen which react, one mole of pressure is lost. We can calculate the pressure of hydrogen in the mixture from the pressure which is lost by using a proportion.
pressure of hydrogen=pressure lost ×21=∆P×2=0.04×2=0.08 atm
7. Dilution Problems
Occasionally stoichiometry problems involve the mixing of two solutions. Before you can do any stoichiometry calculations, you must figure out the result of the dilution. Do this with
MbeforeVbefore = MafterVafter It is helpful to do these problems in two steps. First do the dilution to determine the reactant concentrations; then do the stoichiometric calculation to see what results.
Dilution only -- a common multiple choice question
What is the concentration of potassium ions in the solution which results from mixing 200 mL of 0.050 M K2SO4 with 300 mL of 0.100 M K3PO4?
K+=2×200500×0.050+3×300500×0.100=0.04+0.18=0.22M
Dilution plus reaction
Calcium phosphate, Ca3(PO4)2, is insoluble in water. What is the calcium concentration which results from mixing 200 mL of 0.600 M CaCl2 with 400 mL 0.150 M K3PO4?
Ca2+=200600×0.600=0.200 M
PO43-=400600×0.150=0.100 M
The reaction is: 3 Ca2+ + 2 PO43–® Ca3(PO4)2. So we lose three calcium atoms for every two phosphate ions.
Ca2+=0.200- 32×0.100=0.050 M
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8. Boyle's and Charles' Laws
Calculations involving Boyle's Law (pressure and volume are inversely proportional) and Charles' Law (volume and temperature are directly proportional) frequently occur in multiple choice questions. To do them you need to remember the equation:
A sample of gas occupies 2.00 L at 1.00 atm. and 127° C. At what temperature will it occupy 1.00 L at 1.50 atm?
1.00×2.00400=1.50×1.00T
T=400×1.50×1.001.00×2.00=300 K
T = 27° C
9. The Kinetic Molecular Theory (KMT)
The KMT is a model which explains the behavior of gases under most circumstances. Gases which follow the KMT are referred to as "ideal gases." It can be summarized in several assumptions.
1. Gases consist of particles (atoms and molecules) in constant, random motion.
2. These particles have negligible volume.
3. Collisions between particles are completely elastic.
4. The attractive forces between the particles are negligible.
5. The average kinetic energy of the particles is proportional to their Kelvin temperature.