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!
- 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.
Stoichiometric Calculations definition
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.
Balancing equations
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 O2H2 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!
Mole to mole calculations
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
- 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
Stoichiometric Calculations: example
Using the balanced equation below, calculate the number of moles of iron (III) chloride (FeCl3) that are produced when 1.75 moles of chlorine gas (Cl2) 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 (Cl2) to our desired species (FeCl3). 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 FeCl3:
$$1.75\,mol\,Cl_2*\frac{2\,mol\,FeCl_3}{3\,mol\,Cl_2}$$
$$1.75\cancel{mol\,{Cl_2}}*\frac{2\,mol\,FeCl_3}{3\cancel{mol\,Cl_2}}$$
$$1.75*\frac{2}{3}\,mol\,FeCl_3=1.17\,mol\,FeCl_3$$
Pretty simple right? The fun part is coming up next: mass to mass calculations.
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
- 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 (H2O):
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 (CO2) are produced when 54.6 g of ethane (C2H6) are combusted with enough oxygen gas (O2) 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 CO2:
$$12.01\frac{g}{mol}+2(16.00\frac{g}{mol})=44.01\frac{g}{mol}$$
Finally, we can multiply the number of moles by the molar mass to get the gram amount:
$$3.64\cancel{\,mol}\,CO_2*44.01\frac{g}{\cancel{mol}}\,CO_2=160.2\,g\,CO_2$$
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