Stoichiometric Calculations

When experimenting, you are probably going to need to convert between the masses of species and the molar amount of species. In this article, we will be walking through the steps of these conversions to help you through this process!

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Jetzt kostenlos anmeldenWhen experimenting, you are probably going to need to convert between the masses of species and the molar amount of species. In this article, we will be walking through the steps of these conversions to help you through this process!

- This article cover
**mass-to-mass**and**mole-to-mole**conversions. - First, we will define what a
**stoichiometric calculation**is. - Next, we will learn how to
**balance equations**. - Then, we will learn the
**steps**for**mole to mole conversions**. - After that, we will learn the
**steps**for**mass to mass conversions**. - Lastly, we will
**summarize**these conversion steps.

**Stoichiometry **is the relationship between the quantity of reactants and products in a chemical reaction

**Stoichiometric coefficients **are the numbers in front of a species in a chemical equation. These coefficients tell us the ratio between reactants and products

So what does this all mean? Well first, let's look a general reaction:

$$ 2A + B \rightarrow C + 3D$$

Here we have two reactants (A and B) and two products (C and D). The numbers in front are the stoichiometric coefficients.

These coefficients tell us that is takes 2 moles of A and 1 mole of B to form 1 mol of C and 3 moles of D. When we do stoichiometric calculations, we use these ratios to calculate for different values, such as how many moles of C are produced when 3 moles of A are consumed in the reaction.

Stoichiometric calculations are reliant on stoichiometric coefficients, so we always have to make sure our equations are balanced.

When an equation is balanced, the number of moles of each element are equal on both sides.

For example, here is the balanced equation for the formation of water:

$$2H_{2\,(g)} + O_{2\,(g)} \rightarrow 2H_2O_{(l)}$$

To count the number of moles, we multiply each element's subscript (little number) by their coefficient, so we have:

Hydrogen (left): 2(2)=4 Hydrogen (right) 2(2)=4

Oxygen (left) 1(2)=2 Oxygen (right) 2(1)=2

The number on the left is the coefficient, while the number on the right is the subscript

Now let's work on balancing an equation together

Balance the following equation:

$$NaBr + Ca(OH)_2 \rightarrow CaBr_2 + NaOH$$

Our first step is to count the number of moles of each element on both sides. When elements are in brackets, like (OH) here, we multiply the subscript by *each *element. Therefore, (OH)_{2}, is the same as O_{2}H_{2} in terms of numbers.

Let's start with the left side:

Sodium (Na): 1 Bromine (Br): 1 Calcium (Ca): 1 Oxygen (O): 2 Hydrogen (H): 2

Now for the right:

Sodium (Na): 1 Bromine (Br): 2 Calcium (Ca): 1 Oxygen (O): 1 Hydrogen (H):1

So we have three species that aren't balanced: Br, O, and H

Firstly, we can multiply our sodium hydroxide (NaOH) by 2, so that our oxygen and hydrogen atoms are now balanced:

$$NaBr + Ca(OH)_2 \rightarrow CaBr_2 + 2NaOH$$

Now we have a new problem, our sodium atoms are unbalanced! So for our last step, we can multiply the sodium bromide (NaBr) compound by 2, so that our bromine and sodium atoms are all balanced.

$$2NaBr+ Ca(OH)_2 \rightarrow CaBr_2 + 2NaOH$$

When making calculations, always make sure your equation is balanced!

As you saw earlier, the units for Chemical Equations are in moles, so convert from moles of one species to another isn't two difficult.

The basic steps are:

**Check**if the equation is**balanced**- If
**unbalanced**, balance the equation first

- If
**Multiply**your amount of moles by the**stoichiometric ratio**

So what do I mean by stoichiometric ratio? Well let's walk through an example together to see

*Using the balanced equation below, calculate the number of moles of iron (III) chloride (FeCl _{3}) that are produced when 1.75 moles of chlorine gas (Cl_{2}) are reacted with enough iron (Fe).*

$$2 Fe + 3Cl_2 \rightarrow 2FeCl_3$$

So our first step has been done for us since the equation is already balanced, so onto step 2!

The **stoichiometric ratio **is simply the ratio of our starting species (Cl_{2}) to our desired species (FeCl_{3}). This ratio is given to us by the coefficients.

As long as you make sure your units cancel, this process works to convert from reactants to products, from products to reactants, and from product to product/reactant to reactant.

We multiply our amount of chlorine by this ratio, so that the moles of chlorine cancel, leaving us with moles of FeCl_{3}:

$$1.75\,mol\,Cl_2*\frac{2\,mol\,FeCl_3}{3\,mol\,Cl_2}$$

$$1.75\cancel{mol\,{Cl_2}}*\frac{2\,mol\,FeCl_3}{3\cancel{mol\,Cl_2}}$$

$$1.75*\frac{2}{3}\,mol\,FeCl_3=1.17\,mol\,FeCl_3$$

Pretty simple right? The fun part is coming up next: mass to mass calculations.

Since Chemical Equations are in units of moles, we actually have to convert to moles *first, *then to mass.

The basic steps are:

**Check**if the equation is**balanced**- If
**unbalanced**, balance it first

- If
**Convert**from**mass**of the species to**moles**of the species- Use the
**stoichiometric ratio**to convert from moles of one species to another - Convert from moles back to mass

The way we convert from mass to moles (and vice versa) is by using the **molar mass **of the species:

The **molar mass **of a species is the total mass of an element or compound per 1 mole

When you look at the Periodic Table, the mass listed is the **atomic mass**. The **molar mass of a compound **is the sum of the atomic masses of each element in that compound multiplied by their subscripts.

For example, here's how to calculate the molar mass of water (H_{2}O):

Atomic mass (H): 1.01 g/mol

Atomic mass (O): 16.00 g/mol

$$2(1.01\frac{g}{mol})+16.00\frac{g}{mol}=18.02\frac{g}{mol}$$

Now that we know how to calculate molar mass, let's work on a problem:

Using the balanced equation below, calculate how many grams of Carbon dioxide (CO_{2}) are produced when 54.6 g of ethane (C_{2}H_{6}) are combusted with enough oxygen gas (O_{2}) to react

$$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6 H_2O$$

- Atomic mass (C): 12.01 g/mol Atomic mass (H): 1.01 g/mol
- Atomic mass (O): 16.00 g/mol

For our first step, we need to convert grams of ethane to moles. To do this, we first need to calculate the compound's molar mass:

$$2(12.01\frac{g}{mol})+6(1.01\frac{g}{mol})=30.08\frac{g}{mol}$$

Next, we divide the mass by the molar mass, so that the grams units cancel, and we are left in units of moles:

$$\frac{54.6\cancel{\,g}\,C_2H_6}{\frac{30.08\cancel{\,g}\,C_2H_6}{mol}}=1.82\,mol\,C_2H_6$$

When we divide by a fraction, that is the same as multiplying by the reciprocal (inverse). So when we divide by n g/mol, it is the same as multiplying by mol/ n g

Next, we multiply by the ratio of ethane to Carbon dioxide:

$$1.82\cancel{\,mol\,C_2H_6}*\frac{4\,mol\,CO_2}{2\cancel{\,mol\,C_2H_6}}=3.64\,mol\,CO_2$$

Now we need to convert back to moles, to do this, we multiply by the molar mass. But first, we need to calculate the molar mass of CO_{2}:

$$12.01\frac{g}{mol}+2(16.00\frac{g}{mol})=44.01\frac{g}{mol}$$

Finally, we can multiply the number of moles by the molar mass to get the gram amount:

$$3.64\cancel{\,mol}\,CO_2*44.01\frac{g}{\cancel{mol}}\,CO_2=160.2\,g\,CO_2$$

**The limiting reactant**

In our calculations, we have only been given the amount of one reactant, instead of both. In these example, we assume that we have enough of our second reactant, but that isn't always the case.

In reactions, there is often a **limiting reactant**, which is consumed first. This reactant "limits" how much product can be made.

To find the limiting reactant, you solve for the yield (product made) for each reactant. Whichever one produces less is limiting.

Now that we've learned how to do these calculations, here is a summary of these steps.

**For mole to mole calculations:**

**For mass to mass calculations:**

**Stoichiometry**is the relationship between the quantity of reactants and products in a chemical reaction**Stoichiometric coefficients**are the numbers in front of a species in a chemical equation. These coefficients tell us the ratio between reactants and products- The steps for mole to mole conversions are:
- Check if the equation is balanced
- If unbalanced, balance the equation first

- Multiply your amount of moles by the
**stoichiometric ratio**

- Check if the equation is balanced
- The steps for mass to mass conversions are
- Check if the equation is balanced
- If unbalanced, balance it first

- Convert from mass of the species to moles of the species by dividing by the molar mass
- Use the stoichiometric ratio to convert from moles of one species to another
- Convert from moles back to mass by multiplying by the molar mass

- Check if the equation is balanced
- The
**molar mass**of a species is the total mass of an element or compound per 1 mole- The
**molar mass of a compound**is the sum of the atomic masses of each element in that compound multiplied by their subscripts.

- The

*always *make sure your equation is balanced. Stoichiometric equations rely on balanced chemical equation.

Mole ratios allow us to convert from the moles of one species to another

The molar mass allows us to convert between mass and the molar amount.

**atomic mass**. The **molar mass of a compound **is the sum the atomic masses of each element in that compound multiplied by their subscripts.

What is stoichiometry?

**Stoichiometry **is the relationship between the quantity of reactants and products in a chemical reaction

What are stoichiometric coefficients?

**Stoichiometric coefficients **are the numbers in front of a species in a chemical equation. These coefficients tell us the ratio between reactants and products

What is the first step in stoichiometric calculations?

Balance the equation

Balance the following equation:

$$CaBr_2 + KOH \rightarrow Ca(OH)_2 + KBr$$

$$CaBr_2 + KOH \rightarrow Ca(OH)_2 + KBr$$

What are the basic steps for mole-to-mole calculations?

- Check if the equation is balanced
- If unbalanced, balance the equation first

- Multiply your amount of moles by the
**stoichiometric ratio**

Calculate the amount of moles of ammonia (NH_{3}) formed when 1.2 moles of nitrogen gas (N_{2}) react with enough hydrogen

$$N_2 + H_2 \rightarrow NH_3$$

2.4 mol

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