Mass

Unified Atomic Mass Unit (u)

• The unified atomic mass unit ($\mathrm{u}$) (or dalton ($\mathrm{Da}$))
  • 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 $m_{\mathrm{u}}=\frac{1}{12}m{(}^{12}\text{C})=1\text{u}=1.66053906892(52)\times 10^{-27}\text{kg}$

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.

of an isotope of an element

• atomic mass ($m_{\mathrm{a}}$ or $m$)
  • “Rest mass of an atom in its ground state. The commonly used unit is the unified atomic mass unit.” (IUPAC)
  • units:
    • $\mathrm{kg}$ (the SI unit)
    • $\mathrm{g}$
    • $\mathrm{u}$
  • examples:
    • $m({}_{}^{}{}_{}^{12}\mathrm{C})=12\mathsf{u}$ (the atomic mass of an carbon-12 atom is $12\mathsf{u}$)
    • $m({}_{}^{}{}_{}^{13}\mathrm{C})=13.0033548378\mathsf{u}$ (the atomic mass of an carbon-13 atom is $13.0033548378\mathsf{u}$)
• The relative isotopic mass (of a particular isotope of an element)
  • dimensionless
  • $\frac{m}{1\mathrm{u}}$, where $m$ is the atomic mass of the isotope (in $\mathrm{u}$)
  • (example: the relative isotopic mass of ${}_{}^{}{}_{}^{13}\mathrm{C}$ is $\frac{13.0033548378\mathrm{u}}{1\mathrm{u}}=13.0033548378$)

of an element

• The average atomic mass (or just atomic mass) of an element
  • $m(E)={\sum }_i\left(m_i\cdot f_i\right)$
    • $m(E)$ 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 ${}_{}^{}{}_{}^{12}\mathrm{C}$ and ${}_{}^{}{}_{}^{13}\mathrm{C}$ are $98.93\%$ and $1.07\%$ respectively.
    • The average atomic mass of carbon is $m(\mathrm{C})=\left(m({}_{}^{}{}_{}^{12}\mathrm{C})\times \text{NA}({}_{}^{}{}_{}^{12}\mathrm{C})\right)+\left(m({}_{}^{}{}_{}^{13}\mathrm{C})\times \text{NA}({}_{}^{}{}_{}^{13}\mathrm{C})\right)=12.011\mathsf{u}$.
• relative atomic mass (or atomic weight (deprecated)) (of an element)
  • dimensionless
  • $A_r(E)$
  • ”The ratio of the average mass of the atom to the unified atomic mass unit.” (IUPAC)
  • “An atomic weight (relative atomic mass) of an element from a specified source is the ratio of the average mass per atom of the element to 1/12 of the mass of an atom of 12C.” (Atomic Weights of the Elements 1979)
• standard atomic weight ($A_r^{\circ }(E)$)
  • “the weighted arithmetic mean of the relative isotopic masses of all isotopes of that element weighted by each isotope’s abundance on Earth” (Wikipedia)
  • Fourteen chemical elements are defined as an interval (e.g. $A_r^{\circ }(\mathrm{H})=[1.00784,1.00811]$)
  • abridged atomic weight (the standard atomic weight with numbers reduced to five significant figures)
  • IUPAC Periodic Table of the Isotopes
  • https://ciaaw.org/abridged-atomic-weights.htm
  • https://en.wikipedia.org/wiki/Template:Standard_atomic_weight_of_the_elements

of a compound

• molecular mass (for molecular compound, or formula mass for ionic compound)

from Wikipedia:

“The molecular mass (for molecular compounds) and formula mass (for non-molecular compounds, such as ionic salts) are commonly used as synonyms of molar mass, as the numerical values are identical (for all practical purposes), differing only in units (dalton vs. g/mol or kg/kmol). However, the most authoritative sources define it differently. The difference is that molecular mass is the mass of one specific particle or molecule (a microscopic quantity), while the molar mass is an average over many particles or molecules (a macroscopic quantity). “

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

• molar mass ($M$) (unit: $g/mol$)
  • The molar mass of an element is the mass per mole of its atoms.
  • The molar mass of a molecular compound (or ionic compound) is the mass per mole of its molecules (or its formula units)
    • $M={\sum }_i\left(a_i\times M_i\right)$, 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 $g/mol$)
    • Examples:
      • The molar mass of $\mathrm{H}{}_2\mathrm{O}$ is $2\times 1.008g/mol+15.999g/mol=18.015g/mol$.
      • The molar mass of $\mathrm{CO}{}_2$ is $12.011g/mol+2\times 15.999g/mol=44.01g/mol$.
  • $M=\frac{m}{N_A}$, 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 $\mathrm{Ag}$, calculate the mass (in $\mathrm{g}$) of $\mathrm{Ag}$. Answer: The molar mass of $\mathrm{Ag}$ is $107.87g/mol$, so, by the formula $m=n\times M$, the mass of $0.750\mathsf{mol}$ of $\mathrm{Ag}$ is $m=0.750\mathsf{mol}\times 107.87g/mol=80.9\mathsf{g}$.

Converting Grams to Moles

• Given $25.0\mathsf{g}$ of $\mathrm{H}{}_2\mathrm{O}$, calculate the number of moles of $\mathrm{H}{}_2\mathrm{O}$.
  • Answer: The molar mass of $\mathrm{H}{}_2\mathrm{O}$ is $18.015g/mol$, so, by the formula $n=\frac{m}{M}$, the number of moles of $25.0\mathsf{g}$ of $\mathrm{H}{}_2\mathrm{O}$ is $n=\frac{25.0\mathsf{g}}{18.015g/mol}=1.39\mathsf{mol}$.
• Given $737\mathsf{g}$ of $\mathrm{NaCl}$, calculate the number of moles of $\mathrm{NaCl}$.
  • Answer: The molar mass of $\mathrm{NaCl}$ 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 $\mathrm{NaCl}$ 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
• $\text{Element}=\mathrm{K}$ - The element is potassium
• $P=22.0\%$ - The percentage by mass of $\mathrm{K}$ in the sample,
• $\text{Compound}=\mathrm{KCl}$ - The compound is potassium chloride

Solution:

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

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 $\mathrm{CaCO}{}_3$ and $\mathrm{NaHCO}{}_3$.
• $n(\mathrm{NaHCO}{}_3)=0.010\mathsf{mol}$ - The number of moles of $\mathrm{NaHCO}{}_3$ in the mixture.

Solution:

1. Calculate the molar mass of $\mathrm{NaHCO}{}_3$. $M(\mathrm{NaHCO}{}_3)=84.01g/mol$
2. Calculate the mass of $\mathrm{NaHCO}{}_3$ in the mixture. $m(\mathrm{NaHCO}{}_3)=0.010\mathsf{mol}\times 84.01g/mol=0.8401\mathsf{g}$
3. Calculate the mass of sodium ($\mathrm{Na}$) in $\mathrm{NaHCO}{}_3$. $M(\mathrm{Na})=22.99g/mol$ $m(\mathrm{Na})=0.010\mathsf{mol}\times 22.99g/mol=0.2299\mathsf{g}$
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:

• $m(\text{Mixture})=1.6\mathsf{g}$
• $n(\mathrm{CaCO}{}_3)=0.0133\mathsf{mol}$ of $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NO}{}_2$ (amino acid glycine)
• $M(\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NO}{}_2)=75.07g/mol$

Solution:

1. Calculate the mass of $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NO}{}_2$ in the mixture. $m(\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NO}{}_2)=0.0133\mathsf{mol}\times 75.07g/mol=0.998\mathsf{g}$
2. Determine the mass percent of $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NO}{}_2$ in the mixture. $\text{Mass Percent}=\frac{0.998\mathsf{g}}{1.6\mathsf{g}}\times 100=\boxed{62.4\%}$

Calculating Masses in Reactions

• $2\mathrm{Fe}+3\mathrm{Cl}{}_2\to 2\mathrm{FeCl}{}_3$, 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, $2\mathrm{Fe}+3\mathrm{Cl}{}_2\to 2\mathrm{FeCl}{}_3$)
  2. Calculate the molar mass of the product
     • $M(\mathrm{Fe})=55.85g/mol$
     • $M(\mathrm{Cl}{}_3)=3\times 35.5g/mol$
     • $M(\mathrm{FeCl}{}_3)=M(\mathrm{Fe})+M(\mathrm{Cl}{}_3)=162.2g/mol$
  3. Calculate the number of moles of the limiting reactant (in this example, iron)
     • $n(\mathrm{Fe})=\frac{m(\mathrm{Fe})}{M(\mathrm{Fe})}=\frac{2.24\mathsf{g}}{55.85g/mol}=0.0401\mathsf{mol}$
  4. Calculate the mass of the product that can be produced
     • $m(\mathrm{FeCl}{}_3)=n(\mathrm{Fe})\times M(\mathrm{FeCl}{}_3)=0.0401\mathsf{mol}\times 162.2g/mol=6.50\mathsf{g}$
• 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, $m(\text{product})=n(\text{limiting reactant})\times M(\text{product})$