How to Perform Mole-Mole Conversions from Balanced Equations

How to Perform Mole-Mole Conversions from Balanced Equations

Chemistry Workbook For Dummies, 2nd Edition

You can balance a chemical equation by adjusting the coefficients that precede reactant and product compounds within the equation. After you’ve got a balanced equation, you can use the coefficients to build mole-mole conversion factors. These kinds of mole conversion factors tell you how much of any given product you get by reacting any given amount of reactant.

Consider the following balanced equation for generating ammonia from nitrogen and hydrogen gases:

Industrial chemists around the globe perform this reaction, humorlessly fixating on how much ammonia product they’ll end up with at the end of the day. In any event, how are chemists to judge how closely their reactions have approached completion? The heart of the answer lies in a balanced equation and the mole-mole conversion factors that spring from it.

For every mole of nitrogen reactant, a chemist expects 2 moles of ammonia product. Similarly, for every 3 moles of hydrogen reactant, the chemist expects 2 moles of ammonia product. These expectations are based on the coefficients of the balanced equation and are expressed as mole-mole conversion factors as shown in the figure.

For example, say you want to calculate how many moles of ammonia can be expected from the reaction of 278 mol of N2 gas.

To solve this problem, you begin with your known quantity, the 278 mol of nitrogen that’s to be reacted. You multiply that quantity by the mole-mole conversion factor that relates moles of nitrogen to moles of ammonia. You write the conversion factor so that mol NH3 is on top and mol N2 is on the bottom. That way, the mol N2 units cancel, leaving you with the desired units, mol NH3. The numbers you put in front of the units for the conversion factor come directly from the coefficients in the balanced chemical equation.

So, you expect 556 mol of ammonia from the reaction.

Try another example. One source of hydrogen gas is the electrolysis of water, in which electricity is passed through water to break hydrogen-oxygen bonds, yielding hydrogen and oxygen gases:

Based on this equation, solve these three problems:

How many moles of hydrogen gas result from the electrolysis of 78.4 mol of water?

Keep in mind that the number you use for moles is the coefficient for the compound taken from the balanced chemical reaction:

You find that the electrolysis reaction produces 78.4 moles of hydrogen gas.

How many moles of water are required to produce 905 mol of hydrogen?

To find the moles of water produced, you again apply a mole-mole conversion:

This shows you that the reaction produces 905 moles of water.

Running the electrolysis reaction in reverse constitutes the combustion of hydrogen. How many moles of oxygen are required to combust 84.6 mol of hydrogen?

Finally, you run the electrolysis reaction in reverse:

This shows that 42.3 moles of oxygen are required to combust 84.6 moles of hydrogen.

www.dummies.com

8.2: Mole-to-Mole Conversions

Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.

Consider the simple chemical equation

[2H_2 + O_2 → 2H_2O]

The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as

[4H_2 + 2O_2 → 4H_2O]

The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as

[22H_2 + 11O_2 → 22H_2O]

because 22:11:22 also reduces to 2:1:2.

Suppose we want to use larger numbers. Consider the following coefficients:

[12.044 times 10^<23>; H_2 + 6.022 times 10^<23>; O_2 → 12.044 times 10^<23>; H_2O]

These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 10 23 is 1 mol, while 12.044 × 10 23 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as

[2 ;mol; H_2 + 1; mol; O_2 → 2 ;mol; H_2O]

We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is

[2H_2 + O_2 → 2H_2O]

Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”

By the same token, the ratios we constructed to describe molecules reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:

We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry.

See also:  How To Use

How many moles of oxygen react with hydrogen to produce 27.6 mol of H 2O?

Solution

Unbalanced: H 2 + O 2 → H 2O

Balanced: 2H 2 + O 2 → 2H 2O

Prepare a concept map and use the proper conversion factor.

To produce 27.6 mol of H 2O, 13.8 mol of O 2 react.

Steps for Problem Solving How many moles of oxygen react with hydrogen to produce 27.6 mol of H 2O?
Find a balanced equation that describes the reaction
Identify the «given»information and what the problem is asking you to «find.» Given: moles H2O Find: moles oxygen
List other known quantities 1 mol O2 = 2 mol H2O
Cancel units and calculate.
Think about your result. Since each mole of oxygen produces twice as many moles of water, it makes sense that the produced amount is greater than the reactant amount

Example (PageIndex<2>)

How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?

Solution

Balanced: N2 + 3H22NH3

Prepare a concept map and use the proper conversion factor.

The reaction of (4.20 : text) of hydrogen with excess nitrogen produces (2.80 : text) of ammonia

Steps for Problem Solving How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen
Find a balanced equation that describes the reaction
Identify the «given»information and what the problem is asking you to «find.»
List other known quantities 3 mol H2 = 2 mol NH3
Cancel units and calculate.
Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.

Exercise (PageIndex<1>)

  1. Given the following balanced chemical equation,[ce]how many moles of H2O can be formed if 0.0652 mol of C5H12 were to react?
  2. Balance the following unbalanced equation and determine how many moles of H2O are produced when 1.65 mol of NH3 react.[ce nonumber]

Answer a: 3.14 mol H2O Answer b: 4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol H2O

Summary

  • The balanced chemical reaction can be used to determine molar relationships between substances.

Contributions & Attributions

This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:

chem.libretexts.org

Mass-Mole Calculations (n=m/M) Chemistry Tutorial


Key Concepts

1 mole of a pure substance has a mass equal to its molecular mass (1) expressed in grams.

This is known as the molar mass, M, and has the units g mol -1 (grams per mole of substance)

The relationship between molar mass, mass and moles can be expressed as a mathematical equation as shown below:

g mol -1 = g ÷ mol

molar mass = mass ÷ moles

where:
M = molar mass of the pure substance (measured in g mol -1 )
m = mass of the pure substance (measured in grams, g)
n = amount of the pure substance (measured in moles, mol)

This mathematical equation can be rearranged to give the following:

n = m ÷ M
moles = mass ÷ molar mass

m = n × M
mass = moles × molar mass

To calculate the moles of pure substance: n = m ÷ M

To calculate mass of pure substance: m = n × M

To calculate molar mass of pure substance: M = m ÷ n

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Calculating the Mass of a Pure Substance (m=nM)

1 mole of a pure substance is defined as having a mass in grams equal to its relative molecular mass.
This quantity is known as the molar mass (symbol M).
So, mass of 1 mole of a pure substance = relative molecular mass in grams
And, mass of 1 mole of a pure substance = molar mass of the pure substance (g mol -1 )
Or, mass of 1 mole = M (g mol -1 )

The table below gives the mass of 1 mole of a number of common pure substances:

name molecular formula relative molecular mass molar mass
(g mol -1 )
mass of 1 mole
(g)
helium gas He 4.003 4.003 g mol -1 4.003 g
oxygen gas O2 2 × 16.00 = 32.00 32.00 g mol -1 32.00 g
carbon dioxide gas CO2 12.01 + (2 × 16.00) = 44.01 44.01 g mol -1 44.01 g
liquid water H2O (2 × 1.008) + 16.00 = 18.016 18.016 g mol -1 18.016 g

From the table we see that 1 mole of water has a mass of 18.016 grams, which isn’t very much (about the mass of water in a couple of small ice-cubes you’d make in your family freezer).

But what if you had 10 moles of water? What would be the mass of 10 moles of water?
If 1 mole of water has a mass of 18.016 g, then 10 moles of water must have ten times more mass:
mass of 10 moles of water = 10 × mass of 1 mole of water
mass of 10 moles of water = 10 × 18.016 = 180.16 g (about the mass of water you could put in a small glass)

So, if we only had ½ mole of water, what mass of water would we have?
If 1 mole of water has a mass of 18.016 g, then ½ mole of water must have ½ the mass:
mass of ½ mole of water = ½ × mass of 1 mole of water
mass of ½ mole of water = ½ × 18.016 = 9.008 g

In both of the examples above, we can calculate the mass of water in grams by multiplying the moles of water by the mass of 1 mole of water in grams:

mass water = moles of water × mass of 1 mole water

because the mass of 1 mole of water in grams is known as its molar mass, we can write:

mass water = moles of water × molar mass of water

The table below compares the mass of different amounts of water in moles and the data is graphed on the right:

mass of water
(g)
= moles of water
(mol)
× molar mass of water
(g mol -1 )
= 0.00 × 18.016
9.008 = 0.50 × 18.016
18.016 = 1.00 × 18.016
27.024 = 1.50 × 18.016
180.16 = 10.00 × 18.016
270.24 = 15.00 × 18.016


From the data in the table we can generalise and say that for any pure substance the mass of substance in grams is equal to the moles of substance multiplied by the mass of 1 mole of the substance:

mass = moles × mass of 1 mole

and since mass of 1 mole of a substance (in grams) = molar mass (in grams per mole)

mass (g) = moles × molar mass (g mol -1 )

The graph above shows a straight line that passes through the origin (0,0) so the equation for the line is:
y = slope × x
where y is mass of water (g)
and x is moles of water (mol)
and slope (gradient) of the line
= vertical rise ÷ horizontal run
=270.24g÷15mol=18.016 g mol -1
= molar mass of water (g mol -1 )
So the equation for this line is:
mass (H2O) = molar mass (H2O) × moles (H2O)
In general:
mass (g) = molar mass (g mol -1 ) × moles (mol)

Follow these steps to calculate the mass of a pure substance given the amount of substance in moles:

Extract the data from the question:

mass = m = ? (units are grams)
moles = n = write down what you are told in the question
molar mass = M = write down what you are told in the question (units are g mol -1 )
(you may need to calculate this using the molecular formula of the pure substance and a Periodic Table)

Check the units for consistency and convert if necessary:

The amount of substance must be in moles (mol) !
If amount is given in millimoles (mmol), divide it by 1,000 to give the amount in moles (mol).
If amount is given in micromoles (μmol), divide it by 1,000,000 to give an amount in moles (mol).
If amount is given in kilomoles (kmol), multiply it by 1,000 to give an amount in moles (mol).

Write the mathematical equation (mathematical formula):

Substitute in the values and solve the equation to find the value of mass, m, in grams (g).

Do you know this?

Play the game now!

Calculating the Moles of a Pure Substance (n=m/M)

In the discussion above, we discovered that we could calculate the mass of a pure substance using the moles and molar mass of the substance:

mass (g) = moles (mol) × molar mass (g mol -1 )

How would we calculate the moles of pure substance if we knew the mass of the substance?

(a) We could use some algebra: divide both sides of the equation by the molar mass:

mass = moles × molar mass
molar mass molar mass

moles = mass ÷ molar mass

(b) We could use some logic:

we know the mass with units of grams (g)
we know the molar mass with units of grams per mole (g mol -1 )
we need to find moles with units of mole (mol)

By inspection of units we see that dividing molar mass by mass will give us a quantitiy in units of «mol -1 «

molar mass/mass = g mol -1 / g = mol -1

If we turn this upside down (in mathematical terms, take the reciprocal) we get a quantity with units of «mol» which is what we want:

mass/molar mass = g / g mol -1 = mol

moles = mass ÷ molar mass

Follow these steps to calculate the amount of pure substance in moles given the mass of substance:

Extract the data from the question:

mass = m = write down what you are told in the question
moles = n = ? (units are mol)
molar mass = M = write down what you are told in the question (units are g mol -1 )
(you may need to calculate this using the molecular formula of the pure substance and a Periodic Table)

Check the units for consistency and convert if necessary:

Mass must be in grams !
If mass is given in milligrams (mg), divide it by 1,000 to give the mass in grams (g).
If mass is given in micrograms (μg), divide it by 1,000,000 to give a mass in grams (g).
If mass is given in kilograms (kg), multiply it by 1,000 to give a mass in grams (g).

Write the mathematical equation (mathematical formula):

Substitute in the values and solve the equation to find moles of substances (mol).

Do you understand this?

Take the test now!

Calculating the Molar Mass of a Pure Substance (M=m/n)

What if you knew the amount of a pure substance in moles and its mass?
Could you calculate its molar mass?

Recall that mass = moles × molar mass or m = n × M

(a) We could use some algebra: divide both sides of the equation by the moles:

mass = moles × molar mass
moles moles

molar mass = mass ÷ moles

(b) We could use some logic:

By inspection of units we see that dividing the mass in grams by the amount in moles we arrive at a quantity with the units grams per mole (g mol -1 ) which are the units for molar mass.

Therefore, molar mass (g mol -1 ) = mass (g) ÷ moles (mol)

or you can write

Follow these steps to calculate the molar mass of a pure substance given the amount of substance in moles and the mass of substance:

Extract the data from the question:

mass = m = write down what you are told in the question
moles = n = write down what you are told in the question
molar mass = M = ? (units are g mol -1 )

Check the units for consistency and convert if necessary:

Mass must be in grams (g)!
Amount, moles, must be in moles (mol)!

Write the mathematical equation (mathematical formula):

Substitute in the values and solve the equation to find the molar mass of the substance in grams per mole.

Can you apply this?

Take the exam now!

Worked Examples of Calculating Mass, Moles, Molar Mass


Worked Example: mass = moles × molar mass (m=n×M)

Calculate the mass of 0.25 moles of water, H2O.

Extract the data from the question:

moles = n = 0.25 mol
molar mass = M = (2 × 1.008) + 16.00 = 18.016 g mol -1 (Calculated using the periodic table)
mass = m = ? g

Check the data for consistency:

Is the amount of water in moles (mol)? Yes.
We do not need to convert this.

Write the mathematical equation (mathematical formula):

Substitute the values into the equation and solve for mass (g):

Worked Example: moles = mass &divide molar mass (n=m/M)

Calculate the amount of oxygen gas, O2, in moles present in 124.5 g of oxygen gas.

Extract the data from the question:

mass = m = 124.5 g
molar mass = M = 2 × 16.00 =32.00 g mol -1 (Calculated using the periodic table)
moles = n = ? mol

Check the data for consistency:

Is the mass of oxygen gas in grams (g)? Yes.
We do not need to convert this.

Write the mathematical equation (mathematical formula):

Substitute the values into the equation and solve to find moles of oxygen gas:

Worked Example: molar mass = mass ÷ moles (M=m/n)

Calculate the molar mass of a pure substance if 1.75 moles of the substance has a mass of 29.79 g.

Extract the data from the question:

mass = m = 29.79 g
moles = n = 1.75 mol

Check the data for consistency:

Is the mass of in grams (g)? Yes. We do not need to convert this.
Is the amount of substance in moles (mol)? Yes. We do not need to convert this.

Write the equation:

Substitute the values into the equation and solve for molar mass:

molar mass = M = 29.79 ÷ 1.75 = 17.02 g mol -1

Can you apply this?

Do the drill now!

Problem Solving Using Moles, Mass, and Molar Mass

The Problem: Calcium carbonate, CaCO3, is an important industrial chemical. Chris the Chemist has an impure sample of calcium carbonate. The mass of the impure sample is 0.1250 kg and it is composed of 87.00% (by mass) calcium carbonate. Before Chris can use this calcium carbonate in a chemical reaction, Chris needs to know the amount, in moles, of calcium carbonate present in this sample.

Calculate the amount of calcium carbonate in moles present in this impure sample of calcium carbonate.

Solving the Problem using the StoPGoPS model for problem solving:

Calculate the amount of calcium carbonate in moles

n(CaCO3) = moles of calcium carbonate = ? mol

PAUSE! What chemical principle will you need to apply?

Apply stoichoimetry (n = m ÷ M)

What information (data) have you been given?

mass of sample = m(sample) = 0.1250 kg

Step 1: Write the mathematical equation to calculate moles of calcium carbonate:

Step 2: Calculate the mass of calcium carbonate in the sample in kilograms (kg).

mass of calcium carbonate (kg) = 87.00% of mass of sample (kg)
m(CaCO3) = ( 87.00/100) × m(sample)

Step 3: Convert the mass of calcium carbonate in kilograms (kg) to mass in grams (g)

Step 4: Calculate the molar mass of calcium carbonate (use Periodic Table to find molar mass of each element):

Step 5: Substitute the values for m(CaCO3) in grams and M(CaCO3) in g mol -1 into the mathematical equation and solve for n (mol)

GO! Step 1: Write the mathematical equation to calculate moles of calcium carbonate:

Step 2: Calculate the mass of calcium carbonate in the sample in kilograms (kg).

mass of calcium carbonate (kg) = 87.00% of mass of sample (kg)
m(CaCO3) = ( 87.00/100) × m(sample)
m(CaCO3) = ( 87.00/100) × 0.1250 kg = 0.10875 kg

Step 3: Convert the mass of calcium carbonate in kilograms (kg) to mass in grams (g)

m(CaCO3) in grams = m(CaCO3) in kg × 1000 g/kg
m(CaCO3) in grams = 0.10875 kg × 1000 g/kg 108.75 g

Step 4: Calculate the molar mass of calcium carbonate (use Periodic Table to find molar mass of each element):

molar mass = M(CaCO3) = M(Ca) + M(C) + [3 × M(O)]
M(CaCO3) = 40.08 + 12.01 + [3 × 16.00] = 40.08 + 12.01 + 48.00 = 100.09 g mol -1

Step 5: Substitute the values for m(CaCO3) in grams and M(CaCO3) in g mol -1 into the mathematical equation and solve for n (mol)

PAUSE! Have you answered the question that was asked?

Yes, we have calculated the moles of calcium carbonate in the sample.

Is your solution to the question reasonable?

Let’s work backwards to see if the moles of calcium carbonate we have calculated will give us the correct mass for the sample.
Roughly calculate mass of CaCO3 in 1.087 mol (≈ 1 mol):
m(CaCO3) = n × M = 1 × (40 + 12 + 3 × 16) = 100 g
Roughly calculate the mass of sample if 87% of its mass is due to CaCO3:
m(CaCO3) = 87/100 × m(sample)
m(sample) = 100/87 × m(CaCO3) = 100/87 × 100 = 115 g = 0.115 kg

Since this approximate value for the mass of the sample is about the same as the mass of sample given in the question, we are reasonably confident that our answer is correct.

STOP! How many moles of calcium carbonate are present in the sample?

(1) Molecular mass is also known as molecular weight, formula weight or formula mass

www.ausetute.com.au

Share:
No comments

Добавить комментарий

Your e-mail will not be published. All fields are required.

×
Recommend
Adblock
detector