Skip to content
RMIT University Library - Learning Lab

The mole - exercises

 

\[\require{mhchem}\]

Worked Examples

  1. How many acetaminophen molecules are present in \(0.25\textrm{mol}\) of acetaminophen?
    From the mole concept and the Avogadro’s number, we know that one mole of acetaminophen contains \(6.022\times10^{23}\) acetaminophen molecules. Therefore how many acetaminophen molecules will be in \(0.25\textrm{mol}\) of acetaminophen: \[\begin{align*} \frac{6.022\times10^{23}\textrm{acetaminophen molecules}}{1\textrm{mol}}\times0.25\textrm{mol} & =1.5\times10^{23}\textrm{acetaminophen molecules} \end{align*}\]

  2. Glucose has the formula \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}\). Calculate the molar mass of glucose. The molar masses of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) are given below.
    \(\ce{C}=12.01\textrm{g}/\textrm{mol},\ce{H}=1.01\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\)
    Glucose contains six atoms of \(\ce{C}\), twelve atoms of \(\ce{H}\) and six atoms of \(\ce{O}\):
    \[\begin{align*} \left(6\times12.01\textrm{g}/\textrm{mol}\right)+\left(12\times1.01\textrm{g}/\textrm{mol}\right)+\left(6\times16.00\textrm{g}/\textrm{mol}\right) & =180.1\textrm{g}/\textrm{mol} \end{align*}\]

  3. Aspirin has the formula \(\ce{C}_{9}\ce{H}_{8}\ce{O}_{4}\). Calculate the mass of aspirin in grams present in \(0.65\textrm{mol}\) of aspirin.The molar masses of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) are given below.
    \(\ce{C}=12.01\textrm{g}/\textrm{mol},\ce{H}=1.01\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\)
    The amount of aspirin and molar mass of aspirin (molar mass can be determined from the given information) are given in the question. You need to find the mass of aspirin in \(0.65\) moles. The first step is to calculate the molar mass of aspirin from the given atomic molar masses. \[ \left(9\times12.01\textrm{g}/\textrm{mol}\right)+\left(8\times1.01\textrm{g}/\textrm{mol}\right)+\left(4\times16.00\textrm{g}/\textrm{mol}\right)=180.1\textrm{g}/\textrm{mol} \] The next step is to rearrange the equation to make \(m\) the subject: \[\begin{align*} M & =\frac{m}{n}\\ m & =M\times n \end{align*}\] The final step is to substitute the given values with units to the equation and solve the equation: \[\begin{align*} m & =M\times n\\ m & =180.1\textrm{g}/\textrm{mol}\times0.65\textrm{mol}\\ m & =117\textrm{g} \end{align*}\]

  4. Calculate the amount of molecular oxygen \(\left(\ce{O}_{2}\right)\) present in \(160.0\textrm{g}\) of oxygen. The molar mass of atomic \(\ce{O}\) is \(16.00\textrm{g}/\textrm{mol}\).
    The molar mass of molecular oxygen: \[ 16.00\textrm{g}/\textrm{mol}\times2=32.00\textrm{g}/\textrm{mol} \] The moles of molecular oxygen: \[\begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{160.0\textrm{g}}{32.00\textrm{g}/\textrm{mol}}\\ n & =5.000\textrm{mol} \end{align*}\]

  5. How many moles of \(\ce{C}\), \(\ce{O}\) and \(\ce{H}\) are present in \(1.5\textrm{mol}\) of glucose? The chemical formula of glucose is \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}\).
    From the chemical formula of glucose, we know that one mole of glucose contains six moles of carbon, twelve moles of hydrogen and six moles of oxygen. We can use this relationship to calculate the moles of \(\ce{C}\), \(\ce{O}\) and \(\ce{H}\) present in \(1.5\textrm{mol}\) of glucose as follows:
    Number of \(\ce{C}\) moles in \(1.5\) moles of glucose: \[ \frac{6\textrm{mol }\ce{C}}{1\textrm{mol}\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}}\times1.5\textrm{mol}=9.0\textrm{mol} \] Number of \(\ce{H}\) moles in \(1.5\) moles of glucose: \[ \frac{12\textrm{mol}\ce{H}}{1\textrm{mol}\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}}\times1.5\textrm{mol}=18\textrm{mol} \] Number of \(\ce{O}\) moles in \(1.5\) moles of glucose: \[ \frac{6\textrm{mol}\ce{O}}{1\textrm{mol}\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}}\times1.5\textrm{mol}=9.0\textrm{mol} \]

  6. How many grams of carbon are present in \(1.8\textrm{g}\) of glucose. The chemical formula of glucose is \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}\). The atomic molar masses of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) are given below.
    \(\ce{C}=12.01\textrm{g}/\textrm{mol},\ce{H}=1.01\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\)
    You are asked to calculate the mass of carbon and are given the mass of glucose and atomic molar masses of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\). With the given data you can calculate the number of moles of glucose present in \(1.8\textrm{g}\) of glucose using \(M=\frac{m}{n}\). Then you can use the resulting value to calculate the number of moles of carbon present in \(1.8\textrm{g}\) of glucose. Once you have the number of carbon moles, multiply it by the atomic molar mass of carbon to get the grams.

  • Step 1 - Calculate the molar mass of glucose: \[ \left(12.01\textrm{g}/\textrm{mol}\times6\right)+\left(1.01\textrm{g}/\textrm{mol}\times12\right)+\left(16.00\textrm{g}/\textrm{mol}\times6\right)=180.1\textrm{g}/\textrm{mol} \]

  • Step 2 - Calculate the number of moles of glucose present in \(1.8\textrm{g}\) of glucose: \[\begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{1.8\textrm{g}}{180.1\textrm{g}/\textrm{mol}}\\ n & =0.010\mathscr{\textrm{mol}} \end{align*}\]

  • Step 3 - Calculate the number of moles of carbon present in \(0.01\textrm{mol}\) of glucose: \[ \frac{6\textrm{mol}\ce{C}}{1\textrm{mol}\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}}\times0.010\textrm{mol}\ce{C}_{2}\ce{H}_{12}\ce{O}_{6}=0.060\textrm{mol}\ce{C} \]

  • Step 4 - Convert the number of moles of carbon into grams: \[\begin{align*} M & =\frac{m}{n}\\ m & =n\times M\\ m & =0.060\textrm{mol}\times12.01\textrm{g}/\textrm{mol}\\ m & =0.72\textrm{g} \end{align*}\]

Exercises

  1. How many hydrogen atoms are present in \(1.5\textrm{mol}\) of hydrogen?

    2. Answer

    \(\frac{6.022\times10^{23}\textrm{atoms}}{1\textrm{mol}}\times1.5\textrm{mol}=9.0\times10^{23}\textrm{atoms}\)

  2. Caffeine is one of the main ingredients in coffee and tea that acts as a central nervous system stimulant. The chemical formula of caffeine is \(\ce{C}_{8}\ce{H}_{10}\ce{N}_{4}\ce{O}_{2}\).

  1. What is the molar mass of caffeine? Atomic mass of \(\ce{C}=12.01\textrm{amu},\ce{H}=1.01\textrm{amu},\ce{O}=16.00\textrm{amu},\ce{N}=14.01\textrm{amu}\).
  2. How many molecules of caffeine are present in \(970.95\textrm{g}\) of a caffeine sample?

    3. Answer

    a. \(194.1\textrm{g}/\textrm{mol}\)

    b. \(5.002\textrm{mol}\)

  1. Calculate the mass in grams of a \(1.2\textrm{mol}\) sample of \(\ce{CO}_{2}\). Atomic molar mass of \(\ce{C}\) and \(\ce{O}\) are given below.
    \(\ce{C}=12.01\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\).

    2. Answer

    \(53\textrm{g}\)

  2. A balloon filled with helium gas contains \(1.505\times10^{26}\) atoms of helium. What is the mass of helium present inside the balloon? The atomic molar mass of \(\ce{He}=4.00\textrm{g}/\textrm{mol}\).

    3. Answer

    \(1000\textrm{g}\)

  3. Ribose is a simple sugar present in RNA, the molecule involved in protein synthesis in living organisms. Ribose has the chemical formula \(\ce{C}_{5}\ce{H}_{10}\ce{O}_{5}\). How many moles of \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) are present in \(0.8\textrm{mol}\) of ribose sugar?

    4. Answer

    \(\ce{C}=4\textrm{mol},\ce{H}=8\textrm{mol},\ce{O}=4\textrm{mol}\)

  4. Calcium phosphate, whose chemical formula is \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\), is an essential mineral present in the bones of many living organisms. Calculate the number of formula units of calcium phosphate present in \(31.02\textrm{g}\) of calcium phosphate sample.
    Atomic molar mass of \(\ce{Ca}=40.08\textrm{g}/\textrm{mol},\ce{P}=30.97\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\).

    5. Answer

    \(6.022\times10^{22}\) formula units

  5. \(\ce{CFC}-12\) is one of the Chlorofluorocarbons, which has the chemical formula \(\ce{CCl}_{2}\ce{F}_{2}\). Chlorofluorocarbons have been used as refrigerants for many years. How many grams of chlorine are present in \(12.9\textrm{g}\) of \(\ce{CFC}-12\) sample? Molar mass of \(\ce{CFC}-12\) is \(120.91\textrm{g}/\textrm{mol}\) and \(\ce{Cl}=35.45\textrm{g}/\textrm{mol}\).

    6. Answer

    \(7.56\textrm{g}\)

  6. Keratin is a fibrous protein found in our skin, hair and nails. It has the chemical formula \(\ce{C}_{28}\ce{H}_{48}\ce{N}_{2}\ce{O}_{32}\ce{S}_{4}\). A sample of keratin contains \(0.71\textrm{g}\) of \(\ce{N}\). What is the mass of \(\ce{O}\) present in the sample? Molar mass of \(\ce{N}=14.01\textrm{g}/\textrm{mol},\ce{O}=16.00\textrm{g}/\textrm{mol}\).

    7. Answer

    \(13\textrm{g}\)

Keywords: 
Chemistry
Chemistry exercises
The mole
Measurements