Stoichiometry with Moles - Learnfaster.calearnfaster.ca/wp-content/uploads/2015/05/General-Stoichiometry.pdf · Stoichiometry is a four step process: ... Stoichiometry using solutions

© Copyright Pavel Sedach 2013, All Rights Reserved, Page 1 of 3

Stoichiometry with Moles

Stoichiometry is a four step process: 1. Balance the chemical reaction. 2. Convert all reagent quantities (grams of solid, L and mol/L of liquid and L or atm of gas) to moles 3. Use a molar ratio to find the limiting reagent and the reagent asked for reaction and then convert the

amount of the limiting reagent into a product. 4. Convert moles of the needed reagent to the quantity asked (grams of solid, L and mol/L of liquid and L

or atm of gas) Consider the analogy of 5 bicycle tires and 4 frames used in making complete bicycles. Each bicycle uses 2 tires and 1 frame. Some of you can already see that we will run out of tires before we run out of frames. But how did we do that calculation?

We took 5 wheels

(2 wheels per bicycle)= 2.5 bicycles and we took

4 frames

(1 frame per bicycle) to get 4 bicycles.

The number of bicycles we make is limited by the number of wheels. This is the essence of chemical stoichiometry – let's phrase it as a chemistry problem: I have 5 wheels and 4 frames, given the following equation: 2 wheels + 1 frame 1 bicycle How many bicycles can I make? Your first step is to write down the relevant variables underneath the equation:

2 wheels + 1 frame 1 bicycle Quantity 5 wheels 4 frames Then we convert using a molar ratio:

5 wheels ×1 bicycle

2 wheels= 2.5 bicycles

and

4 frames ×1 bicycle

1 frame= 4 bicycles

In this way we find the 'limiting reagent', the wheels.


The above example is the same as saying: How many moles of C2H6(g) will I form if I have 5 moles of H2(g) and 4 moles of C2H2(g)?

C2H2(g) + H2(g) → C2H6(g)

First we balance the reaction and list our variables Elements are always balanced last (changing the coefficient for 𝐇𝟐(𝐠) will not affect the number of carbons)

C2H2(g) + 2 H2(g) → C2H6(g)

4 moles 5 moles You may be inclined to say the C2H2(g) is the limiting reagent because it has less moles, however you are

forgetting you need to convert it into the moles of the product using a molar ratio:

5 moles H2(g) ×1 C2H6(g)

2 H2(g)= 2.5 moles C2H6(g)

and

4 moles C2H2(g) ×1 C2H6(g)

1 C2H2(g)= 4 moles C2H6(g)

*Often times you will see molar ratio written as 1 mol C2H6(g)

1 mol C2H2(g), notice the mol on top and bottom. They cancel so

I never write mol in molar ratios. I just use the coefficients directly. Similar to the bicycle question that came before, our limiting reagent was the wheels (H2(g)) and we know we

can make a maximum of 2.5 bicycles (moles of C2H6(g))

*We can make 2.5 moles because a mol is 6.022 × 1023 things. We can’t actually make 0.5 bicycles but if we had moles of bicycles, it would be possible.


Percent Yield

Consider the analogy of 5 bicycle tires and 4 frames used to make bicycles. Each bicycle uses 2 tires and so we know we will run out of tires before we run out of frames. But now, let’s add an extra factor. You get a really bad call while assembling one of the bicycles, and your hand slipped and was cut on a gear. In a blinding rage you beat the bicycle with your wrench and the frame and wheels got bent out of whack. Well, this time the ‘theoretical yield’ is the same: 2 wheels + 1 frame --> 1 bicycle 5 wheels 4 frames 2 bicycles are possible but we only made one (after attending anger management class and building the second bicycle more carefully). In this way, our yield is 50% (1 out of 2). In chemistry things work the same way: What is the percent yield of C2H6(g) if 1 mol of C2H6(g) is formed when 5 moles of H2(g) and 3 moles of

C2H2(g) react?

C2H2(g) + 2 H2(g) → C2H6(g)

3 moles 5 moles We can theoretically produce 2.5 moles C2H6(g) as before but now we must calculate the theoretical yield:

actual yield

theoretical yield=

1 mol C2H6(g)

2.5 mol C2H6(g)= 0.4 = 40% yield

*I never write × 100% as it is redundant. A decimal IS a percentage. Notice that when we calculate “percent yield” the words tell you what to do!!! A percent is a fraction out of something – so, you need to make a ratio. Yield means what you can make. So a percent yield is a fraction out of what you can actually make. In this way there’s no formula to memorize – just use the English language to help you. Think about it – why is the yield not always 100%? If you’ve ever baked you know some of the ingredients line the pan when you remove the dough, so there already is some loss. And let’s say you burned the cake after leaving it too long in the oven – that charred part isn’t cake – it’s a new product of a chemical reaction with oxygen! Fascinating, but sadly, not edible – and also, not a contributor to your yield.


Stoichiometry using masses Stoichiometry is a four step process:

1. Balance the chemical reaction. 2. Convert all reagent quantities (grams of solid, L and mol/L of liquid and L or atm of gas) to moles 3. Use a molar ratio to find the limiting reagent and the reagent asked for reaction and then convert the


or atm of gas)

Example:

If 5 g of CH4(g) react with 16 g of O2(g), what will the yield of CO2(g) be in grams?

The above is a combustion reaction. Generally in combustion reactions, elements in the compound react as

follows:

C reacts with O2(g) to make CO2(g) (if oxygen is limiting, CO(g) can be produced)

H reacts with O2(g) to make H2O(g)

S reacts with O2(g) to make SO2(g) (this product can vary)

N reacts with O2(g) to make NO2(g) (this product can vary)

Using the tips above, we write the equation as follows:

CH4(g) + O2(g) → CO2(g) + H2O(g)

After balancing (remember we balance the element, O2(g), last):

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

When you first learn stoichiometry you complete a lot of redundant work. As a chemist gains experience they

usually just put the relevant information in a table as the conversions are pretty straightforward:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Molecular Weight (g/mol) 16.05 32.00 44.01

grams 5 16 ?

moles


*Molecular weight calculations:

CH4(g) = 12.01 + 4(1.01) =16.05 grams

mole

O2(g) = 2(16.00) =32.00 grams

mole

CO2(g) = 12.01 + 2(16.00) =44.01 grams

mole

*Gram to mole calculations:

5 grams of CH4(g) ×1mol

16.05 g= 0.312 moles of CH4(g)

16 grams of O2(g) ×1mol

32.00 g= 0.500 moles of O2(g)

Now we do a mole to mole calculation to see what is limiting:

0.312 mol CH4(g) ×1 CO2(g)

1 CH4(g)= 0.312 mol CO2(g)

0.500 moles of O2(g) ×1 CO2(g)

2 O2(g)= 0.250 mol CO2(g)

We now know O2(g) limits the production of CO2(g). We can then convert the moles of CO2(g) created by the

limiting reagent into grams as follows:

0.250 mol CO2(g) ×44.01 grams

mole= 11.0025 grams CO2(g)

That’s it.

Let’s ask another question – if the actual yield of 𝐂𝐎𝟐(𝐠) was 9.5 g, what was the percent yield?

actual yield

theoretical yield=

9.5 grams CO2(g)

11.0025 grams CO2(g)= 0.863 = 86.3% yield


Dilution Whenever we are diluting a substance, we add more solvent. This means the number of moles stays the same:

moles = moles

mol

L× L =

mol

L× L

This gives

C1V1 = C2V2

Where an initial concentration can be diluted to a final one.

Example: How much 1.00 M H2SO4(aq) do I need in order to make 5.00 L of a 0.15 mol/L sulfuric acid solution?

C1V1 = C2V2,1.00 mol

L× V1 =

0.15 mol

L5.00 L

V1 = 0.750 L

This means we needed to dilute 750 mL of Fluid to 5000 mL. We can ask a further question- how much water

did we add to dilute the solution?

5000 mL – 750 mL = 4250 mL

All of these are possible questions.


Stoichiometry using solutions Example 1: How many g of 𝐌𝐠𝐂𝐥𝟐(𝐚𝐪) are produced when excess 𝐌𝐠(𝐬) fully reacts with 5 L of 10 mol/L

𝐇𝐂𝐥(𝐚𝐪)?

*this reaction requires you to know that acids generally react with metals to produce the metal salt and

hydrogen gas.

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

Molecular Weight (g/mol) 36.46 95.21

grams ?

moles

moles/liter 10 mol/L

liters 5 L

We start by manipulating the quantity of the reagent(s) we know most about into the moles of reagent we are

interested in:

5.00 L HCl(aq) ×10 mol

L= 50 moles HCl(aq)

*When we start these types of calculations, we start them with a physical quantity – liters, moles, grams – this

is the actual amount of a substance we can hold/manipulate. Afterwards, we always multiply them by a

conversion ratio – mol/L, g/mol, etc. to convert them into the quantity we want. These ratios can be flipped at

will to give you the correct unit – this is called dimensional analysis because the units cancel. For instance we

can say there are 10 people

room which can be restated as

1 room

10 people. You use this type of math all the time to do room

bookings, cooking, finances, home renovation projects, etc. – you’ve just likely never thought about it.

*Also notice that it was unnecessary to find the molecular weight of HCl(aq) – we only need to find the

number of particles in a specific volume – not their weight.

Next we convert moles of our reagent into the moles of what we are looking for:

50 moles HCl(aq) ×1 MgCl2(aq)

1 HCl(aq)= 50 moles MgCl2(aq)

Then we convert moles of MgCl2(aq) to grams:

50 moles MgCl2(aq) ×95.21 grams

mole= 4760.5 grams MgCl2(aq) = 4.76 kg


Example 2: How many g of 𝐌𝐠(𝐬) are required to fully react with 5 L of 10 mol/L 𝐇𝐂𝐥(𝐚𝐪)?


Molecular Weight (g/mol) 24.31 36.46

grams ? ?

moles

moles/liter 10 mol/L

liters 5 L

*In the above question, we are converting one reactant into another – this just illustrates that you can use

molar ratios backwards and forwards (if you know amount of products you can find amount of reactants and

vice versa)

50 moles HCl(aq) ×1 Mg(s)

1 HCl(aq)= 50 moles Mg(s)

50 moles Mg(s) ×24.31 grams

mole= 1215.5 grams Mg(s) = 1.22 kg

*The examples given above are unrealistic because as the concentration of HCl(aq) and therefore acidity of

the solution decreases the rate of reaction would slow to a crawl if not stop altogether.


Example 3: If I titrated 20 mL of 𝐇𝐂𝐥(𝐚𝐪) to equivalence using 10 mL of 0.10 M NaOH, what was my

concentration of 𝐇𝐂𝐥(𝐚𝐪)?

NaOH(aq) + HCl(aq) → H2O(aq) + NaCl(aq)

*we first write a balanced reaction. In acid base reactions I find it easier to write water as HOH(aq). In this way

I can see that the OH from the sodium hydroxide reacts 1:1 with the H from hydrochloric acid – leading to a

balanced reaction.

NaOH(aq) + HCl(aq) → HOH(aq) + NaCl(aq)

moles

moles/liter 0.10 mol/L

liters 0.010 L 0.020 L

*I removed the grams and molecular weight rows from the table as they are irrelevant to finding moles this

time around.

0.01 L NaOH(aq) ×0.1 mol

L×

1 HCl

1 NaOH×

1

0.020 L=

0.050 mol HCl(aq)

L

Notice that the above calculation condensed steps 2 through 4 into 1 calculation – convert L to moles, moles

to moles and then moles to mol/L. As you get faster at identifying the aim of a stoichiometry problem and

finding the limiting reagent, you too can speed up your calculations this way. In the meantime the 3 steps are

given below:

0.01 L NaOH(aq) ×0.1 mol

L= 0.001 mol of NaOH(aq)

0.001 mol of NaOH(aq) ×1 HCl

1 NaOH= 0.001 mol of HCl(aq)

0.001 mol of HCl(aq) ÷ 0.020 =0.050 mol HCl(aq)

L

These steps lead to the same answer and if you look at the unit cancellation it also works out (the dimensional

analysis).


Stoichiometry using gases Stoichiometry is a four step process:

1. Balance the chemical reaction. 2. Convert all reagent quantities (grams of solid, L and mol/L of liquid and L or atm of gas) to moles 3. Use a molar ratio to find the limiting reagent and the reagent asked for reaction and then convert the


or atm of gas)

Example:

If I react 300 mL of 0.250 mol/L HCl with 3.00 grams of Mg(s), what is my yield of H2(g) in liters of gas at STP?

First, we create a balanced reaction and fill in all the details on each reagent:


3.00 g 300 mL ? L

24.31 g/mol

0.250 mol/L

STP

moles ? mol Then, we convert all known quantities into moles:

5 grams of Mg(s) ×1mol

24.31 g= 0.123 moles of Mg(s)

0.300 L of HCl(aq) ×0.250 mol

L= 0.0750 moles of HCl(aq)

Now we do a mole to mole calculation to see what is limiting:

0.123 mol Mg(s) ×1 H2(g)

1 Mg(s)= 0.123 mol H2(g)

0.0750 mol HCl(aq) ×1 H2(g)

2 HCl(aq)= 0.0375 mol H2(g)

We see that HCl(aq) is the limiting reagent because it limits the production of H2(g).

We know we make 0.0375 mol H2(g) at STP so now we simply plug our unknowns into the ideal gas equation:

PV = nRT

(1 atm)V = (0.0375 mol) (0.08205Latm

molK) (273.15 K)

Rearranging for V we get

V = 0.840 L


*The above question is quite unrealistic. As the acidity of the HCl(aq) solution approaches neutral, because

H(𝑎𝑞)+ is used up, the rate of reaction would go to practically zero and the Mg(s) would stop reacting. For the

purposes of this exercise it is OK to assume all the H(𝑎𝑞)+ is reacted.

Documents

Stoichiometry with Moles - Learnfaster.calearnfaster.ca/wp-content/uploads/2015/05/General-Stoichiometry.pdf · Stoichiometry is a four step process: ... Stoichiometry using solutions