# 4.10: Molecular Weight: Conversions


Learning Objectives
• Apply a molecular weight conversion factor to convert between a molar quantity and a compound mass.
• Apply multiple conversion factors to convert between a compound mass and a particle count.

In the previous section, the molecular weights (MWs) of several compounds were calculated by adding the mass contributions of the constituent elements found within each corresponding compound.  Upon completing each calculation, the resultant solution was then represented as a molar equality, a conversion factor, and a "hidden" conversion factor.  As has been the case with the Avogadro's number, "component within," and atomic weight molar standards, a molecular weight conversion factor can be applied to bring about a desired unit transformation.

## Applying Molecular Weight Conversion Factors in Dimensional Analysis

As stated previously, the quantity containing the unit that is being canceled must be written in the denominator of a conversion factor, in order to bring about a desired unit transformation.  Orienting the applied conversion factor in this way will cause the given unit, which appears in a numerator, to be divided by itself and, therefore, "cancel," since the same unit appears in the denominator of the conversion factor.  Remember that all numerical values that have been discussed in this chapter are associated with two units.  Therefore, in order to achieve complete unit cancelation, a conversion factor that results in the simultaneous elimination of both units must be applied.

For example, use a conversion factor based on the molecular weight of ZnBr2, 225.19 g/mol ZnBr2, to calculate how many grams of ZnBr2 are present in 0.25 moles of ZnBr2.

The reference to a mass unit, "grams," in the given problem, coupled with the knowledge that ZnBr2, zinc bromide, is a compound, indicates that a mass-based molecular weight conversion factor should be applied to solve this problem.  The "/" in the given molecular weight unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor.  Because the word "per" implies a ratio, which is associated with the mathematical operation of division, the word "per" is represented as the fraction bar in a conversion factor.  Any number or unit read before the word "per" becomes the numerator of the conversion, and any number or unit found after the word "per" is written in the denominator.  The value "225.19 g/mol ZnBr2" can, therefore, be written as

$$\dfrac{225.19 {\text{ g }} \ce{ZnBr_2}}{\text{ mol } \ce{ZnBr_2}}$$ or $$\dfrac{\text{ mol } \ce{ZnBr_2}}{225.19 {\text{ g }} \ce{ZnBr_2}}$$

The conversion factor on the left is a direct representation of the given molecular weight, and the second conversion factor is derived by interchanging where each quantity is written, relative to the fraction bar.  However, only the conversion factor on the left will allow for the complete cancelation of the given unit, "moles of ZnBr2," since both of the units that are being canceled must be written in the denominator of the conversion factor that should be applied to solve the given problem.  Therefore,

$${0.25 \; \cancel{\rm{mol} \; \rm{ZnBr_2}}} \times$$ $$\dfrac{225.19 \; \rm{g} \; \rm{ZnBr_2}}{\; \cancel{\rm{mol} \; \rm{ZnBr_2}}}$$ = $${\text {56.2975}}$$ $${\rm{g} \; \rm{ZnBr_2}}$$ ≈ $${\text {56}}$$ $${\rm{g} \; \rm{ZnBr_2}}$$

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator.  When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division.  Finally, the correct number of significant figures should be applied to any calculated quantity.  Since the math involved in dimensional analysis is multiplication and division, the number of significant figures in each number being multiplied or divided must be counted, and the answer must be limited to the lesser count of significant figures.  Neither the given number nor the molecular weight that was utilized above are exact numbers.  As the given number contains fewer significant figures, the final answer should be rounded to two significant digits, as shown above.

Example $$\PageIndex{2}$$

Use a conversion factor based on the molecular weight of copper (I) sulfite, 207.17 g/mol Cu2SO3, to calculate how many moles of copper (I) sulfite are present in 6,200 milligrams of copper (I) sulfite.

Solution

The reference to a mass unit, "milligrams," in the given problem, coupled with the knowledge that copper (I) sulfite, Cu2SO3, is a compound, indicates that a mass-based molecular weight conversion factor should be applied to solve this problem.  The "/" in the given molecular weight unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor.  In particular, the word "per" implies a ratio, which is associated with the mathematical operation of division.  Therefore, the word "per" is written as the fraction bar in a conversion factor.  Any number or unit read before the word "per" becomes the numerator of the conversion, and any number or unit found after the word "per" is written in the denominator.  The value 207.17 g/mol Cu2SO3 can, therefore, be written as

$$\dfrac{207.17 {\text{ g }} \ce{Cu_2_SO_3}}{\text{ mol } \ce{Cu_2_SO_3}}$$ or $$\dfrac{\text{ mol } \ce{Cu_2_SO_3}}{207.17 {\text{ g }} \ce{Cu_2_SO_3}}$$

The conversion factor on the left is a direct representation of the given molecular weight, and the second conversion factor is derived by interchanging where each quantity is written, relative to the fraction bar.  However, neither of these conversion factors can be used to completely eliminate the unit "milligrams of copper (I) sulfite," as the mass unit that is referenced in the problem, "milligrams," and the mass unit included in the molecular weight equality, "grams," do not match.  Therefore, a prefix modifier equality that relates these two units must first be developed and applied.  As discussed in Section 1.8, a unit that contains a prefix modifier, such as "milligrams," can be related back to its base unit, "grams," by replacing the prefix modifier, "milli" (m), with its meaning, "10-3".  The base unit, gram (g), is written with both the prefix modifier and its meaning.  Finally, since their base units (grams) match, and the prefix modifier and its meaning are equivalent, these two quantities can be equated to one another, as shown below.

mg = 10-3 g

While not absolutely necessary, a prefix modifier equality can be simplified by rewriting the numerical value in decimal format, as follows.

mg = 0.001 g

The decimal value can be eliminated by dividing both quantities by 0.001, resulting in the equality shown below.

1,000 mg = g

Before applying one of these relationships to cancel the given unit, the chemical formula for copper (I) sulfite, Cu2SO3, must be added as a secondary unit on each side of the equal sign, as all equalities in this chapter must contain two units.  After incorporating this chemical formula, any of these equalities can be applied to completely cancel the given unit, "milligrams of copper (I) sulfite."  Regardless of which equality is selected, the quantity on the left side of the equal sign becomes the denominator in the first conversion factor that is applied to solve the given problem.  The remaining portion of the equality is written in the numerator in the resultant conversion factor.

The unit that results upon the cancelation of "milligrams of copper (I) sulfite" is "g Cu2SO3," which is not the desired final unit.  As a result, the second molecular weight conversion factor that is shown above must be applied as an additional conversion factor, as shown below.

$${\text {6,200}}$$ $${\cancel{\rm{mg} \; \rm{Cu_2SO_3}}} \times$$ $$\dfrac{ \; \bcancel{\rm{g} \; \rm{Cu_2SO_3}}}{1,000\; \cancel{\rm{mg} \; \rm{Cu_2SO_3}}}$$ × $$\dfrac{ \; \rm{mol} \; \rm{Cu_2SO_3}}{207.17\; \bcancel{\rm{g} \; \rm{Cu_2SO_3}}}$$ = $${\text {0.0299271...}}$$ $${\rm{mol} \; \rm{Cu_2SO_3}}$$

≈ $${\text {0.030}}$$ $${\rm{mol} \; \rm{Cu_2SO_3}}$$

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator.  When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.

Exercise $$\PageIndex{1}$$

Consider a problem that requires a calculation of the mass of molecular nitrogen that contains 7.2153 × 1024 molecules of molecular nitrogen.  The molecular weight of molecular nitrogen is 28.02 g/mol N2

1. Identify the indicator information in the given problem, and state which molar standard is associated with each indicator.
2. Write an equality pattern and/or a conversion factor that corresponds to each of the indicators identified in Part (a).
3. Using dimensional analysis, apply the equalities and/or conversion factors that were developed in Part (b) as conversion factors, in order to bring about the desired unit transformation and solve the given problem.
Indicator Information
The word "mass," coupled with the knowledge that molecular nitrogen, N2, is a compound, indicates that a mass-based molecular weight conversion factor should applied to solve this problem.

Additionally, the word "molecules" indicates that an Avogadro's number equality should be developed and applied to solve this problem.

Equality Patterns and Conversion Factors
Molecular Weight Conversion Factor
The word "mass," coupled with the knowledge that molecular nitrogen, N2, is a compound, indicates that a mass-based molecular weight conversion factor should applied to solve this problem.  The "/" in the given molecular weight unit is read as "per," which implies a ratio.  Therefore, the word "per" is written as the fraction bar in a conversion factor, any number or unit read before the word "per" becomes the numerator of the conversion, and any number or unit found after the word "per" is written in the denominator.  The value 28.02 g/mol N2 can, therefore, be written as

$$\dfrac{28.02 {\text{ g }} \ce{N_2}}{\text{ mol } \ce{N_2}}$$ or $$\dfrac{\text{ mol } \ce{N_2}}{28.02 {\text{ g }} \ce{N_2}}$$

The word "molecules" indicates that an Avogadro's number equality should also be developed.  Furthermore, since "molecules" is an indicator word, this word is inserted as the second unit on the right side of this type of equality.  The chemical name that is referenced in the given statement, "molecular nitrogen," should not be used in this, or any, equality.  Instead, the corresponding chemical formula, "N2," which is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas, is incorporated into both of the secondary unit positions in the equality that is being developed.  The resultant Avogadro's number equality is shown below.

1 mol N2 = 6.02 × 1023 N2 molecules

Because the given chemical information is the name of a covalent molecule, the indicator word "molecules" appropriately corresponds to the chemical formula that is applied in this equality.

Dimensional Analysis
In order to completely eliminate the given unit, "molecules of molecular nitrogen," a conversion factor based on the Avogadro's number equality must be applied first.  Specifically, the quantity on the right side of this equality becomes the denominator in the first conversion factor that is applied to solve the given problem.  While reversing the order of the two units on the right side of an Avogadro's number equality is not absolutely necessary, doing so more clearly illustrates the desired unit cancelation for this particular problem.  The remaining portion of the equality is written in the numerator in the resultant conversion factor.

However, the unit that results upon the cancelation of "molecules of molecular nitrogen" is "mol N2," which is not the desired final unit.  Therefore, the first molecular weight conversion factor that is shown above must be applied as an additional conversion factor, as shown below.

$${7.2153 \times 10^{24} \; \cancel{\rm{molecules} \; \rm{N_2}}} \times$$ $$\dfrac{1 \; \bcancel{\rm{mol} \; \rm{N_2}}}{6.02 \times 10^{23} \; \cancel{\rm{molecules} \; \rm{N_2}}}$$ × $$\dfrac{28.02 \; \rm{g} \; \rm{N_2}}{ \; \bcancel{\rm{mol} \; \rm{N_2}}}$$ = $${\text {335.835...}}$$ $${\rm{g} \; \rm{N_2}}$$ ≈ $${\text {336}}$$ $${\rm{g} \; \rm{N_2}}$$

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator.  When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.

Finally, note that this problem simply required a calculation of a "mass of molecular nitrogen," and no particular unit was specified.  Therefore, expressing the final answer in grams, as shown above, is acceptable.  Using an alternative unit, such as kilograms, would also be reasonable.

4.10: Molecular Weight: Conversions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.