16.19: Stoichiometry Map

Solving problems in chemistry requires choosing the right types of equations and equivalencies to get from what we know to what we want to know. In stoichiometry, we have often complex calculations, with many steps, so we need to organized to make sure we're choosing the right steps. Below, you can find a summarized map of steps for solving various types of stoichiometric problems. For all of these problems, we have some type of measurement or measurements of one substance, but we want to know something about a different substance. We also know or can find the balanced equation that relates these two substances to each other. Notice that no matter what information we start or end with, we need to find the moles! Each of the relationships in this map is explained further below.

Finding moles of A

We can find the moles of A from a wide range of information. You do not need the balanced equation to find the moles of A, but you would need to know the molar mass of the substance!

If we're given the grams of A:

This was covered in previous chapter, but it helps to have a reminder of the steps:

1. Find the chemical formula of A, if it isn't given to you
2. Use the chemical formula to find the molar mass (remember, molar mass is measured in g/mol)
3. Use the molar mass and dimensional analysis to convert the g A to mol A

Example \(\PageIndex{1}\)

Convert 2.34 g of sodium chloride to moles.

Solution

1. the formula for sodium chloride is NaCl
2. Use the Periodic Table to find the molar mass of NaCl.
   • Mass of Na = 22.989769 g/mol
   • Mass of Cl = 35.453 g/mol
   • Add the masses = 22.989769 g/mol + 35.453 g/mol = 58.442769 g/mol
   • Round for proper significant figures = 58.443 g/mol (remember - we're using addition, so we round based on the least decimal places)
3. Use dimensional analysis to perform the conversion

\[2.34 \; \cancel{g\;NaCl} \left( \dfrac{1\; mol\;NaCl}{58.443\;\cancel{g\;NaCl}}\right)= 0.400\; mol\;NaCl\nonumber\]

If we're given the particles of A:

This is review, too! Here's a reminder of your steps:

1. Make sure the information given matches A. For example, if you are given that there are 1.2 * 10²⁴ molecules of CO₂, and the balanced equation includes CO₂, then you'll be able to go to the next step directly. If, instead, you are given that there are 1.2 * 10²⁴ ions from NaCl and the balanced equation is for NaCl, you'll first have to convert your number of ions to your formula units of NaCl.
2. Use Avogadro's number and dimensional analysis to convert from particles to moles, 6.022*10²³ particles = 1 mole of particle.

Example \(\PageIndex{2}\)

If there are 1.2*10²⁴ ions from NaCl, how many moles of NaCl is this?

Solution

1. Since you are given ions, not the full NaCl, you will first need to convert the number of ions to the number of formula units of NaCl. If you recall, each NaCl will break up into one Na⁺ ion and one Cl^- ion. This gives the following relationship:
   • 1 NaCl = 2 ions
   • Using this relationship, we can convert the ions to full NaCl:
   • \[1.2*10^{24} \; \cancel{ions} \left( \dfrac{1\; NaCl}{2\;\cancel{ions}}\right)= 6.0*10^{23}\; NaCl\nonumber\]
2. Convert the formula unit of NaCl to moles
   • \[6.0*10^{23} \; \cancel{formula\;units\;NaCl} \left( \dfrac{1\; mole\;NaCl}{6.022*10^{23}\;\cancel{formula\;units\;NaCl}}\right)= 1.0\; mol\;NaCl\nonumber\]

 

Converting A to moles, if it's a Gas

In the previous 2 examples, the substance we're converting could be a liquid, a gas, or a solid. There are specific, additional calculations we can do if A is a gas. We can only use these if A is a gas! Check the balanced equation - is there a (g) after the substance? This is also review, but there are two different conditions we look at for gases, so be careful to use the right one! Here's our steps:

1. Check if the gas is at STP (standard temperature and pressure). STP is only at a temperature of 273 K (or 0 degrees C) and a pressure of 1 atm. You will be told that the gas is at STP or given both the temperature and pressure. If the gas is at STP, then you can use the molar volume to find the moles of gas. This relationship is given below:
   • 1 mole of a gas = 22.4 L at STP
2. If the gas is not at STP, you'll typically need to use the ideal gas law and solve for the moles of gas (n from the equation). Ideal gas law:
   • PV = nRT

Example \(\PageIndex{3}\)

Convert 85.3 L of a gas at STP to moles.

Solution

This gas is at STP, so we can use the molar volume to convert the volume of the gas to the moles:

\[85.3 \; \cancel{L\;gas} \left( \dfrac{1\; mol\;gas}{22.4\;\cancel{L\;gas}}\right)= 3.81\; mol\;gas\nonumber\]

 

Example \(\PageIndex{4}\)

How many moles of hydrogen gas is there if the sample of gas is at a pressure of 750 mm Hg, a volume of 1.35 L, and a temperature of 45 degrees C?

Solution

This gas is not at STP, so we'll need to use the ideal gas law. Remember, to use the ideal gas law, we need the units to match the units of R, the ideal gas constant. We'll use the following R value:

\[\dfrac{ 0.0821 \;L*atm}{mol*K} \nonumber\]

That means the units need to match this R value.

From our problem:

P = 750 mm Hg, which does not match the pressure units of atm from the R value, so we'll need to convert it before we can continue.

\[750 \; \cancel{mm\; Hg} \left( \dfrac{1\; atm}{760\;\cancel{mm\;Hg}}\right)= 0.99\; atm\nonumber\]

V = 1.35 L, which matches the units of R, so we don't need to convert it

T = 45 degrees C, which does not match, so we'll need to convert to K. Remember, we always need temperature in kelvin to be able to do gas calculations.

45 C + 273.15 = 318 K

We'll also need to use the ideal gas law, PV=nRT, and rearrange to solve for n.

\[n =\dfrac{ PV}{RT} \nonumber\]

Substitute for the values from the problem and the known R value:

\[n =\dfrac{0.99\;atm*1.35\;L}{318K*\dfrac{ 0.0821 \;L*atm}{mol*K}} \nonumber\]

= 0.051 moles of hydrogen gas

Converting A to moles, if it's a Solution

The previous examples all assume you have a direct measurement of the pure substance, such as pure solid NaCl or pure hydrogen gas, but many substances are easier to handle and more common as a mixture, particularly as a solution. With these types of situations, we would need to use the information we have about the solution to find the moles of the component we are interested in. Be careful about these - if your mixture has multiple solutes in it, then you'll need to make sure you focus on the right solute. Common solutions with many solutes in them include foods, beverages, air, and blood. Here's how we approach these problems:

1. Identify if you are working with a solution. There are a variety of hints you might get about this - you won't always be told directly. Some ways to tell:
   • In the balanced equation, solutes that are dissolved in water are marked with (aq) - if the equation is fully labeled, this is a quick way to tell
   • We know or can more easily measure the volume (L or mL) of the solution and its Molarity
   • Rarely - you may know other measurements of the solution, such as the concentration in percent by mass, percent by volume, or percent mass/volume and either the volume or mass of the solution. These require more steps to solve, but are still solvable. We'll focus on Molarity for this course, but do be aware you could encounter other types of information in lab or clinical settings.
2. Once we know the volume and Molarity of a solute in a solution, we can find the moles using the Molarity equation

Example \(\PageIndex{5}\)

You measure out 12.32 mL of 0.1085 M NaOH in lab. How many moles of NaOH is this?

Solution

We are given a volume, 12.32 mL, and a concentration, 0.1085 M NaOH. The concentration is M, which is Molarity, so we can use the Molarity equation.

\[M =\dfrac{moles\;solute}{L\;solution} \nonumber\]

We need to rearrange the equation to solve for moles solute:

M * L solution = moles solute

Notice that this problem requires L of solution, but we know the volume of the solution in mL. That means we'll need to convert the volume to L before we do the problem:

\[12.32 \; \cancel{mL} \left( \dfrac{10^{-3}\;L}{1\;\cancel{mL}}\right)= 0.01232\; L\nonumber\]

Plug in our M and the volume in L to solve for moles solute:

moles solute = 0.1085 M * 0.01232 L = 0.00133672 moles NaOH = 0.001337 moles NaOH (once we round for significant figures)

You can verify the units by remembering that the units of M can also be rewritten as mol/L.

Converting moles A to moles B

We have one, single piece of information that allows us to convert from moles of one compound to moles of another compound - that's our balanced equation! Here's how we do that:

1. Check that your equation is complete and balanced.
   • If you don't know what the formulas are for the products, you'll need to figure that out first.
   • Make sure the equation is balanced - remember, you can only change the coefficients to balance a chemical equation, you cannot change the formulas of the reactants or products.
2. Use the molar ratio from the coefficients in the balanced equation to convert from moles of A to moles of B.

Example \(\PageIndex{6}\)

If 0.351 mol Zn fully reacts with excess aqueous hydrochloric acid, how many moles of ZnCl₂ would be made?

Solution

Since this is a mole to mole conversion (notice we are starting with a different substance than we end with), we'll need to use the balanced equation. That means writing it out. Based on the information given, we can determine a few things:

• Our first reactant is Zn, which is in its element form. Looking at the Periodic Table, or going from the experience of seeing Zinc in lab or lab videos, we can determine that Zn would be a solid.
• Our second reactant is aqueous hydrochloric acid. Based on the name, we know the formula is HCl and that it is (aq)
• One of our products has to be ZnCl₂ since we're told it is "made."
• We'll need to write out the equation to determine what, if anything, is missing.

\[ \ce{Zn} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow  \ce{ZnCl_2} + \ce{unknown} \nonumber\]

When we see this equation, we need to know what is going on to predict the products. Notice that the "H" in HCl from the reactants is replaced by "Zn" in ZnCl₂ for the products. This tells us that we're probably working with a single replacement reaction. That tells us that the unknown missing compound would be hydrogen. We know from previous chapters that when hydrogen is by itself it forms H₂, and it is found as a gas. Our ZnCl₂ is also missing its physical state. We can use our solubility rules to find that it would be soluble in water, so it must be (aq).

This gives the equation:

\[ \ce{Zn} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow  \ce{ZnCl_2\;(aq)} + \ce{H_2} \left( g \right)\nonumber\]

This equation isn't balanced, so we need to balance it:

\[ \ce{Zn} \left( s \right) + \ce{2\;HCl} \left( aq \right) \rightarrow  \ce{ZnCl_2\;(aq)} + \ce{H_2} \left( g \right)\nonumber\]

Now we can use the balanced equation to find our moles of ZnCl₂ formed.

\[0.351 \; \cancel{mol\;Zn} \left( \dfrac{1\;mol\;ZnCl_2}{1\;\cancel{mol\;Zn}}\right)= 0.351\; mol\;ZnCl_2\nonumber\]

Notice that the 1's in this conversion factor come from the coefficients in the balanced equation. These could be any value, based on what they are in the balanced equation. Notice also that we only needed the balanced equation to convert moles to moles. 

Using moles B to find other values for B

Generally, once we have the moles of B, we can simply reverse the processes we did to find the moles of A to convert the moles of B back into grams, particles, pressure, volume, temperature, or Molarity. Be very careful to track what you know so you pick an appropriate process! Make a plan!