Mass

Atomic Mass

• The atomic mass ($m$ or $m_{\text{a}}$) is the mass of an atom
  • Although the SI unit of mass is the $\mathrm{kg}$, atomic mass is often expressed in the non-SI unit of unified atomic mass units (u).
  • Examples:
    • $m(\text{ce}{}^{12}C)=12\mathsf{u}$ (the atomic mass of an carbon-12 atom is $12\mathsf{u}$)
    • $m(\text{ce}{}^{13}C)=13.0033548378\mathsf{u}$ (the atomic mass of an carbon-13 atom is $13.0033548378\mathsf{u}$)

Unified Atomic Mass Unit (u)

• An unified atomic mass unit (u) (or dalton (Da)) is a non-SI unit of mass defined as $\frac{1}{12}$ of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and at rest.
  • The atomic mass constant is a constant defined as $m_u=\frac{1}{12}m{(}^{12}\text{C})=1\text{u}=1.66053906892(52)\times 10^{-27}\text{kg}$

Average Atomic Mass

Because substances are usually not isotopically pure, (i.e., they are a mixture of isotopes of the element) it is convenient to use the average atomic mass (or atomic weight) which is the weighted average of the atomic masses of the naturally occurring isotopes of the element to calculate the atomic mass of given sample of the element. - General formula for calculating the average atomic mass of an element: - $m(E)={\sum }_i\left(m_i\cdot f_i\right)$, where: - $m(E)$ is the average atomic mass of the element $E$ (in $\mathrm{u}$) - $m_i$ is the atomic mass of the $i$-th isotope of the element (in $\mathrm{u}$) - $f_i$ is the natural abundance of the $i$-th isotope of the element (in $\%$) - Example: - The natural abundance of $\text{ce}{}^{12}C$ and $\text{ce}{}^{13}C$ are $98.93\%$ and $1.07\%$ respectively. - The average atomic mass of carbon is $m(\text{ce}C)=\left(m(\text{ce}{}^{12}C)\times \text{NA}(\text{ce}{}^{12}C)\right)+\left(m(\text{ce}{}^{13}C)\times \text{NA}(\text{ce}{}^{13}C)\right)=12.011\mathsf{u}$

Relative Atomic Mass

the meaning of the term relative in this context is that the atomic mass is a dimensionless quantity, as it is the ratio of the mass of an atom to the unified atomic mass unit.

• The relative isotopic mass of a particular isotope of an element is the mass of the isotope relative to the mass of carbon-12.
  • It is defined as $m_i=\frac{m}{m_{\text{C}}}$, where:
    • $m$ is the atomic mass of the isotope (in $\mathrm{u}$)
    • $m_{\text{C}}$ is the atomic mass of $\text{ce}{}^{12}C$ (in $\mathrm{u}$)
    • $m_i$ is the relative isotopic mass of the isotope (dimensionless)
  • Example:
    • The relative isotopic mass of $\text{ce}{}^{13}C$ is $m_{\text{ce}{}^{13}C}=\frac{13.0033548378}{12}=1.083$
• The relative atomic mass ($A_r$) is a dimensionless quantity.

  • $A_r=\frac{\text{average atomic mass of the element in u}}{\frac{1}{12}\times \text{atomic mass of }\text{ce}{}^{12}C}=\frac{m(E)}{1\text{u}}$

  • Example: $A_r(\text{ce}C)=\frac{12.011\mathsf{u}}{1\mathsf{u}}=12.011$

  • The standard atomic weight ($A_r^{\circ }(E)$ for an element $E$) the weighted arithmetic mean of the relative isotopic masses of all isotopes of that element weighted by each isotope’s abundance on Earth.

    • Formula: $A_r^{\circ }(E)={\sum }_i\left(m_i\cdot f_i\right)$, where:
      • $m_i$ is the relative isotopic mass of the $i$-th isotope of the element (dimensionless)
      • $f_i$ is the natural abundance of the $i$-th isotope
    • Examples:
      • (carbon) $A_r^{\circ }(\text{ce}C)=12.011$

Amount of Substance

• The amount of substance (n) of a sample of matter is the number of entities (atoms, molecules, ions, etc.) in the sample.
• Mole (mol) is the SI unit of amount of substance.
  • The Avogadro number is $N_0=6.02214076\times 10^{23}$.
  • $1\mathsf{mol}$ of a substance contains exactly $N_0$ entities.
  • The Avogadro constant $N_{\text{A}}=6.02214076\times 10^{23}{\text{mol}}^{-1}$
  • Approximately one mole is based on the number of atoms in 12 grams of carbon-12.

Molar Mass

• The molar mass ($M$) of an element/compound is the mass of one mole of a substance of that element/compound.
  • The molar mass is expressed in $g/mol$.
• The molar mass of an element is numerically equal to the atomic mass of the element in $\mathrm{u}$.
  • Example:
    • The molar mass of carbon is $12.011g/mol$.
    • The molar mass of oxygen is $15.999g/mol$.
    • The molar mass of hydrogen is $1.008g/mol$.
• The molar mass of a compound is the sum of the molar masses of the elements in the compound.
  • To determine the molar mass of a compound, multiply the molar mass of each element by its subscript in the formula and add the results.
  • $M={\sum }_i\left(a_i\times M_i\right)$, where:
    • $M$ is the molar mass of the compound (in $g/mol$)
    • $a_i$ is the subscript of the $i$-th element in the compound’s chemical formula (which is the number of atoms of that element in one molecule of the compound).
    • $M_i$ is the molar mass of the $i$-th element (in $g/mol$)
  • Example:
    • The molar mass of $\text{ce}H2O$ is $2\times 1.008g/mol+15.999g/mol=18.015g/mol$.
    • The molar mass of $\text{ce}CO2$ is $12.011g/mol+2\times 15.999g/mol=44.01g/mol$.
• The molar mass $M$ of a substance can be calculated by these formulas:
  • $M=\frac{m}{N_A}$, where:
    • $m$ is the mass of the substance (in $\mathrm{g}$)
    • $N_0$ is Avogadro’s number
  • $M=\frac{m}{n}$, where:
    • $m$ is the mass of the substance (in $\mathrm{g}$)
    • $n$ is the number of moles of the substance (in $\mathrm{mol}$)

Converting Moles to Grams

Given $0.750\mathsf{mol}$ of $\text{ce}Ag$, calculate the mass (in $\mathrm{g}$) of $\text{ce}Ag$. Answer: The molar mass of $\text{ce}Ag$ is $107.87g/mol$, so, by the formula $m=n\times M$, the mass of $0.750\mathsf{mol}$ of $\text{ce}Ag$ is $m=0.750\mathsf{mol}\times 107.87g/mol=80.9\mathsf{g}$.

Converting Grams to Moles

• Given $25.0\mathsf{g}$ of $\text{ce}H2O$, calculate the number of moles of $\text{ce}H2O$.
  • Answer: The molar mass of $\text{ce}H2O$ is $18.015g/mol$, so, by the formula $n=\frac{m}{M}$, the number of moles of $25.0\mathsf{g}$ of $\text{ce}H2O$ is $n=\frac{25.0\mathsf{g}}{18.015g/mol}=1.39\mathsf{mol}$.
• Given $737\mathsf{g}$ of $\text{ce}NaCl$, calculate the number of moles of $\text{ce}NaCl$.
  • Answer: The molar mass of $\text{ce}NaCl$ is $(22.99+35.45)g/mol=58.44g/mol$, so, by the formula $n=\frac{m}{M}$, the number of moles of $737\mathsf{g}$ of $\text{ce}NaCl$ is $n=\frac{737\mathsf{g}}{58.44g/mol}=12.6\mathsf{mol}$.

Calculating the mass of a substance in a mixture

Problem: Given:

• $m(\text{Sample})$ - The mass of a sample containing a compound.
• $P$ - The percentage by mass of a specific element in the sample.

Find:

• $M(\text{Compound})$ - The molar mass of the compound
• $m(\text{Element})$ - The mass of the element in the sample.
• $m(\text{Compound})$ - The mass of the compound required to provide the calculated mass of the element in the sample.

Answer:

1. Calculate the molar mass of the compound. See Molar Mass (of a compound)

2. Calculate the mass of the element in the sample. $m(\text{Element})=\frac{P}{100}\times m(\text{Sample})$

3. Determine the mass of the compound required to provide the calculated mass of the element. $m(\text{Compound})=m(\text{Element})\times \frac{M(\text{Compound})}{M(\text{Element})}$

EXAMPLE

• $m\text{(Sample)}=0.450\mathsf{g}$ - The mass of a sample containing a compound
• $\text{Element}=\text{ce}K$ - The element is potassium
• $P=22.0\%$ - The percentage by mass of $\text{ce}K$ in the sample,
• $\text{Compound}=\text{ce}KCl$ - The compound is potassium chloride

Solution:

1. $M(\text{ce}KCl)=74.55g/mol$ is the molar mass of $\text{ce}KCl$
2. $m(\text{ce}K)=\frac{22.0}{100}\times 0.450=0.099\mathsf{g}$ is the mass of $\text{ce}K$ in the sample
3. $m(\text{ce}KCl)=0.099\mathsf{g}\times \frac{74.55g/mol}{39.10g/mol}=\boxed{0.188\mathsf{g}}$ is the mass of $\text{ce}KCl$.

Calculating the Mass Percent of an Element in a Mixture

Problem: Given:

• $m(\text{Mixture})$ - The mass of a mixture containing different compounds.
• $n(\text{Compound})$ - The number of moles of a specific compound in the mixture.

Find:

• $M(\text{Compound})$ - The molar mass of the specific compound.
• $m(\text{Element})$ - The mass of the specific element in the mixture.
• $\text{Mass Percent}$ - The mass percent of the specific element in the mixture.

Answer:

1. Calculate the molar mass of the compound. See Molar Mass (of a compound)

2. Calculate the mass of the compound in the mixture. $m(\text{Compound})=n(\text{Compound})\times M(\text{Compound})$

3. Calculate the mass of the element in the compound. $m(\text{Element})=n(\text{Compound})\times M(\text{Element in Compound})$

4. Determine the mass percent of the element in the mixture. $\text{Mass Percent}=\frac{m(\text{Element})}{m(\text{Mixture})}\times 100$

EXAMPLE

Given:

• $m(\text{Mixture})=1.5\mathsf{g}$ - The mass of a mixture containing $\text{ce}CaCO3$ and $\text{ce}NaHCO3$.
• $n(\text{ce}NaHCO3)=0.010\mathsf{mol}$ - The number of moles of $\text{ce}NaHCO3$ in the mixture.

Solution:

1. Calculate the molar mass of $\text{ce}NaHCO3$. $M(\text{ce}NaHCO3)=84.01g/mol$
2. Calculate the mass of $\text{ce}NaHCO3$ in the mixture. $m(\text{ce}NaHCO3)=0.010\mathsf{mol}\times 84.01g/mol=0.8401\mathsf{g}$
3. Calculate the mass of sodium ($\text{ce}Na$) in $\text{ce}NaHCO3$. $M(\text{ce}Na)=22.99g/mol$ $m(\text{ce}Na)=0.010\mathsf{mol}\times 22.99g/mol=0.2299\mathsf{g}$
4. Determine the mass percent of sodium in the mixture. $\text{Mass Percent}=\frac{0.2299\mathsf{g}}{1.5\mathsf{g}}\times 100=\boxed{15.33\%}$

Calculating the Mass Percent of a Compound in a Mixture

Problem: Given:

• $m(\text{Mixture})$ - The mass of a mixture containing different compounds.
• $n(\text{Compound})$ - The number of moles of a specific compound in the mixture.
• $M(\text{Compound})$ - The molar mass of the specific compound.

Find:

• $m(\text{Compound})$ - The mass of the compound in the mixture.
• $\text{Mass Percent}$ - The mass percent of the compound in the mixture.

Answer:

1. Calculate the mass of the compound in the mixture. $m(\text{Compound})=n(\text{Compound})\times M(\text{Compound})$

2. Determine the mass percent of the compound in the mixture. $\text{Mass Percent}=\frac{m(\text{Compound})}{m(\text{Mixture})}\times 100$

EXAMPLE

Given:

• $m(\text{Mixture})=1.6\mathsf{g}$
• $n(\text{ce}CaCO3)=0.0133\mathsf{mol}$ of $\text{ce}C2H5NO2$ (amino acid glycine)
• $M(\text{ce}C2H5NO2)=75.07g/mol$

Solution:

1. Calculate the mass of $\text{ce}C2H5NO2$ in the mixture. $m(\text{ce}C2H5NO2)=0.0133\mathsf{mol}\times 75.07g/mol=0.998\mathsf{g}$
2. Determine the mass percent of $\text{ce}C2H5NO2$ in the mixture. $\text{Mass Percent}=\frac{0.998\mathsf{g}}{1.6\mathsf{g}}\times 100=\boxed{62.4\%}$

Calculating Masses in Reactions

• $\text{ce}2Fe+3Cl2\to 2FeCl3$, what’s the maximum mass of iron chloride that can be produced from 2.24 g of iron reacts with excess chlorine?
  1. Write the balanced chemical equation (in this example, $\text{ce}2Fe+3Cl2\to 2FeCl3$)
  2. Calculate the molar mass of the product
     • $M(\text{ce}Fe)=55.85g/mol$
     • $M(\text{ce}Cl3)=3\times 35.5g/mol$
     • $M(\text{ce}FeCl3)=M(\text{ce}Fe)+M(\text{ce}Cl3)=162.2g/mol$
  3. Calculate the number of moles of the limiting reactant (in this example, iron)
     • $n(\text{ce}Fe)=\frac{m(\text{ce}Fe)}{M(\text{ce}Fe)}=\frac{2.24\mathsf{g}}{55.85g/mol}=0.0401\mathsf{mol}$
  4. Calculate the mass of the product that can be produced
     • $m(\text{ce}FeCl3)=n(\text{ce}Fe)\times M(\text{ce}FeCl3)=0.0401\mathsf{mol}\times 162.2g/mol=6.50\mathsf{g}$
• In general, given a balanced chemical equation, we want to find the maximum mass of a product that can be produced when a certain mass of a reactant is used, the mass of the product is the product of the number of moles of the limiting reactant and the molar mass of the product, $m(\text{product})=n(\text{limiting reactant})\times M(\text{product})$