2.9: Calculations Determining the Mass, Moles and Number of Particles
- Page ID
- 50437
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- Convert grams of a substance to moles
- Convert moles of a substance to grams
Counting by Measuring
In this section we will learn to "count" the number of particles of a solid substance by measuring its mass. This is building off of section 2.8.2.2, molar mass of a compound and Dimensional analysis (section1B.4), where the molar mass is considered an equivalent statement that allows you to convert from a measured value of mass to a counted value of number of particles. As such, it is like a unit conversion based on a physical constant (section 2B.3.5.3), the molar mass of a compound.
We will use the following symbols
m = mass (g)
n = moles (mol)
Formula Weight is defined as \[fw=\frac{m(g)}{n(mol)} \\ mass = moles \; times \; formula \; weight \]
This formula weight is an equivalence statement that depends on the identity of the chemical and converts between its mass and its number of moles. For water
\[1\; mol H_2O =18 \; grams \\ fw(H_2O) = 18 \frac{g}{mol}\]
\[m=n(fw)\]
and there are thus two conversion factors
\[\underbrace{ \frac{m}{fw}}_{\text{converts mass to mols}}\leftarrow or\rightarrow \underbrace{n(fw)}_{\text{converts mols to mass }} \]
That is, dividing the mass by the formula weight converts it to the number of moles, and multiply the molese by the formula weight converts its moles to mass. The formula weight relates mass to moles the way density relates mass to volume, and it is a constant that for any element or compound that can be determined from the periodic table.
Grams to Moles
Figure \(\PageIndex{1}\) shows the flow chart for converting mass to moles. I
The best way to solve these problems is to set up a conversion factor based on the equivalence statement (Equation 2.9.2), and then multiply by a conversion factor based on this equivalence statement, where you put the mass in the denominator (which is the same as dividing by the molar mass).
When Using Conversion Factors Keep Units in Equations
Example \(\PageIndex{1}\)
How many molecules are in 3.00g of ibuprofen?
Solution
We calculated the molar mass of ibuprofen in section 2.8.2.2.1 to be 206.27 g/mol.
From the molar mass we can generate the equivalence statement:
\[ \textcolor{red} {206.27g\; C_{13} H_{18}O_2 = 1\: mole\; C_{13} H_{18}O_2}\]
We are converting a mass of 3.00 g, which has 3 significant Figures, and so will use one extra for the molar mass, and use 206.3 g/mol. From
\[3.00\cancel{g}\, C_{13} H_{18}O_2 \underbrace{\left ( \frac{\textcolor{red} { mol\, C_{13} H_{18}O_2}}{\textcolor{red} {206.3\,\cancel{g}}} \right )}_\text{converts mass to mol}=\text{0.0145 mol or 14.5mmol}\]
or we can convert to actual molecules, but need a second equivalence statement
mole = 6.02214x1023 molecules ibuprofen
\[3.00\cancel{g}\, C_{13} H_{18}O_2 \underbrace{\left ( \frac{ \textcolor{red} { \cancel{mol}\, C_{13} H_{18}O_2}}{ \textcolor{red} { 206.3\,\cancel{g}}} \right )}_\text{converts mass to mol}\underbrace{\left ( \frac{ \textcolor{blue} { 6.022x10^{23}\, C_{13} H_{18}O_2}}{ \textcolor{blue} {\cancel{mol}}} \right )}_\text{converts moles to molecules}=8.76x10^{21}\text{ molecules of ibuprofen} \]
NOTE: Either Answer is Correct.
Moles to Grams
Figure \(\PageIndex{2}\): shows the flow chart of converting number of particles to mass, but once again, we will use the conversion factor based on the equivalence statement.
The Trick: The Molar Mass is a conversion factor like a Physical Constant, and so make a Equivalence Statement out of it.
Example \(\PageIndex{2}\)
Using the molar mass of 206.27 g/mol, calculate the mass in grams of 10 molecules of ibuprofen?
Solution
To answer this we need two equivalence statements
- 1 mole = 206.27 g ibuprofen
- 1 mole = 6.02214x1023 molecules ibuprofen
Note: 10 molecules is an exact number and has no significant digits Since the molar mass was given with five significant digits, we would not use 6.022x1023 for Avogadro's number because that has 4, and so we have to decide how many to use with Avogadro's number. The rule of thumb (see below) when deciding how many to use is to use one more significant digit than the number you are working with, and so we choose six.
\[10\cancel{molecules}\, C_{13} H_{18}O_2 \underbrace{\left( \frac{\cancel{\textcolor{blue}{mol\, C_{13} H_{18}O_2}}}{ \textcolor{blue}{6.02214\; x10^{23}\cancel{molecules}}} \right )}_\text{converts molecules to moles}\underbrace{\left ( \frac{ \textcolor{red} {206.27g\; C_{13} H_{18}O_2}}{\cancel{ \textcolor{red} {mol}}} \right )}_\text{converts moles to mass}=3.4252x10^{-21}\text{ g of ibuprofen}\]
Exercise \(\PageIndex{1}\)
Answer the following Questions
- How many moles is 12.2 g of Sodium Bicarbonate?
- How many moles is 18.8 g of Ammonium Nitrate?
- How many atoms is 30.8 g of Ammoium Nitrate?
- What is the mass of 2.45 mole Silver Chloride?
- what is the mass of 3.44 mole of benzene (C6H6) ?
- Answer a
-
Sodium Bicarbonate (NaHCO3) = 0.145 mol
- Answer b
-
Ammonium Nitrate (NH4NO3) = 0.235 mol
- Answer c
-
Potassium Perchlorate (ClKO4) = 0.222mol
- Answer d
-
Silver Chloride (AgCl) = 351g
- Answer e
-
Benzene (C6H6) = 269g
Significant Digits in Molar Mass Calculations
The number of significant digits you use in calculating the molar mass depends on your data. The rule of thumb is use one more significant digit than the measured value that determines the number of significant digits in your final answer. There is no need to use more, but if you use less, then the significant digits of the molar mass determines the significant digits of your final answer.
You should redo equation 2.9.5 using a molar mass of 206.27 g/mol and you will see the final answer (to three significant digits) is the same. In making these conversions, you should always use a molar mass with one extra significant digit than your measured number. If you had uses 3.00000g of ibuprofen (6 significant digits), then if possible you should have used seven significant digits for the molar mass. If you had used five significant digits (206.27g/mol) your answer would have 5 significant digits, even though your measured value had seven.
Contributors and Attributions
Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:
- some material adapted from Paul Flowers, Klaus Theopold & Richard Langley via OpenStax (LibreText Copy section 3.1)