10.5: Mole-mass and mass-mass problems (2023)

last change
save as PDF

Side-id: 288482

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!- \!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{ span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart }{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\ norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm {span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\ mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{ \ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{ \unicode[.8,0]{x212B}}\)

Learning Objectives

Conversion of the mass or moles of one substance to the mass or moles of another substance in a chemical reaction.

We have found that a balanced chemical equation is balanced with respect to both moles and atoms or molecules. We have used balanced equations to set up ratios, now in moles, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with how many moles of reactant B. We can extend this technique even further. Remember that we can use molar mass to relate a molar amount to a mass amount. We can use this ability to answer stoichiometric questions using the masses of a given substance in addition to moles. We do this in the following order:

Collectively, these conversions are referred to as molar mass calculations.

As an example, consider the balanced chemical equation

\[\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \nonumber \nonumber \]

If we have 3.59 mol of Fe₂Ö₃, how many grams of SO₃can react with it? Using the molar mass calculation sequence, we can determine the required mass of SO₃in two steps. First, we construct the corresponding molar ratio determined from the balanced chemical equation to calculate the number of moles of SO₃required. Then use the molar mass of SO₃As a conversion factor, we determine the mass of this number of moles of SO₃hat.

The first step is similar to the exercises we performed in Section 6.4. As usual, we start with the amount we got:

\[\mathrm{3.59\:mol\:Fe_2O_3\times\dfrac{3\:mol\:SO_3}{1\:mol\:Fe_2O_3}=10.77\:mol\:SO_3}\notal\ nonumber\]

The Mole Fairy₂Ö₃Units cancel and moles of SO remain₃Unit. Now let's take that answer and convert it to grams of SO₃, using the molar mass of SO₃as a conversion factor:

\[\mathrm{10.77\: mol\: SO_3\times\dfrac{80.07\: g\: SO_3}{1\: mol\: SO_3}=862.4\: g\: SO_3} \nonumber \nonumber \]

Our final answer is expressed in three important numbers. In a two-step process, we find 862 g of SO₃reacts with 3.59 mol Fe₂Ö₃. Many problems of this nature can be answered in this way.

The same two-step problem can also be solved in a single line instead of two separate steps, like this:

Example \(\PageIndex{1}\)

How many grams of CO₂occurs when 2.09 moles of HCl react according to this balanced chemical equation?

\[\ce{CaCO_3+2HCl\højrepil CaCl_2+CO_2+H_2O}\nonummer\]

Solution

Our strategy will be to convert moles of HCl to moles of CO₂and then from moles of CO₂to grams of CO₂. We need the molar mass of CO₂44.01 g/mol. Carrying out these two transformations in a single line yields 46.0 g of CO₂:

The molar ratio of CO₂and HCl comes from the balanced chemical equation.

A practice

How many grams of glucose (C₆H₁₂Ö₆) occurs when 17.3 mol H₂O is implemented according to this balanced chemical equation?

\[\ce{6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2} \nonummer \]

Response: \(\mathrm{17.3\:mol\:H_2O\times\dfrac{1\:mol\:C_6H_{12}O_6}{6\:mol\:H_2O}\times\dfrac{180.18\: g\:C_6H_{12}O_6}{1\:mol\:C_6H_{12}O_6}=520\:g\:C_6H_{12}O_6}\)

It is a small step from molar mass calculations to mass-to-mass calculations. If we start with a known mass of a substance in a chemical reaction (rather than a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass to moles by using the molar mass of the substance as the conversion factor. Then—and only then—do we use the balanced chemical equation to construct a conversion factor to convert that quantity into moles of another substance, which in turn can be converted into an equivalent mass. The process runs sequentially as follows:

This three-part process can be performed in three discrete steps or combined into a single calculation that includes three conversion factors. The following example illustrates both techniques.

Example \(\PageIndex{2}\): Chlorination of carbon

Methane can react with elemental chlorine to form carbon tetrachloride (\(\ce{CCl4}\)). The balanced chemical equation is as follows:

\[\ce{CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl} \nonummer \]

How many grams of HCl are formed by reacting 100.0 g of \(\ce{CH4}\)?

Solution

First, let's go through the problem step by step. We start by converting the mass of \(\ce{CH4}\) to moles \(\ce{CH4}\) by finding the molar mass of \(\ce{CH4}\) (16.05 g / mol) use the conversion factor:

\(\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}=6.231\: mol\: CH_4}\)

Note that we have inverted the molar mass so that the gram units cancel and we get an answer in moles. Next, we use the balanced chemical equation to find the ratio of moles \(\ce{CH4}\) to moles of HCl and convert our first result to moles of HCl:

\(\mathrm{6.231\: mol\: CH_4\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}=24.92\: mol\: HCl}\)

Finally, we use the molar mass of HCl (36.46 g/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:

\(\mathrm{24,92\: mol\: HCl\times\dfrac{36,46\: g\: HCl}{1\: mol\: HCl}=908,5\: g\: HCl}\)

At each step, we limited the answer to the correct number of significant figures. If desired, we can perform all three conversions in a single line:

\(\mathrm{100.0\: g\: CH_4\ gange \dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}\ gange \dfrac{4\: mol\: HCl}{1\ : mol\: CH_4}\times\dfrac{36,46\: g\: HCl}{1\: mol\: HCl}=908,7\: g\: HCl}\)

This final answer is slightly different from our first answer because only the final answer is limited to the correct number of significant digits. In the first answer, we limited each intermediate size to the correct number of significant digits. As you can see, both answers are essentially the same.

Exercise \(\PageIndex{2}\): Oxidation of propanal

The oxidation of propanal (CH₃CH₂GIVE) to propionic acid (CH₃CH₂COOH) has the following chemical equation:

CH₃CH₂FOR + 2K CZK₂Cr₂Ö₇→ CH₃CH₂COOH + other products

How many grams of propionic acid are formed by reacting 135.8 g of K?₂Cr₂Ö₇?

Response: \(\mathrm{135.8\: g\: K_2Cr_2O_7\times\dfrac{1\: mol\: K_2Cr_2O_7}{294.20\: g\: K_2Cr_2O_7}\times\dfrac{1\: mol\: CH_3CH_2COOH : mol\: K_2Cr_2O_7}\times\dfrac{74.09\: g\: CH_3CH_2COOH}{1\: mol\: CH_3CH_2COOH}=17.10\: g\: CH_3CH_2COOH}\)

For Your Health: The Synthesis of Taxol

Taxol is a potent anticancer drug originally derived from the Pacific yew tree (Taxus brevifolia). As you can see from the attached figure, taxol is a very complicated molecule with the molecular formula \(\ce{C47H51NO14}\). Isolating taxol from its natural source presents certain challenges, primarily because the Pacific yew tree is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related yew species also produce taxol in small amounts, there is great interest in synthesizing this complex molecule in the laboratory.

In 1994, after 20 years of work, two research groups announced the full laboratory synthesis of taxol. However, each synthesis required over 30 separate chemical reactions with an overall efficiency of less than 0.05%. To put this into perspective, to get a single dose of 300mg of Taxol, you need to start with 600g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year in one dose, almost 16,000 kg (over 17 tons) of starting material must be converted into taxol. Taxol is also used to treat breast cancer, which is diagnosed in 200,000 women in the United States each year. This only increases the amount of raw material needed.

Clearly, there is great interest in increasing the overall efficiency of taxol synthesis. Improved synthesis will not only be easier, but will also produce less waste, allowing more people to benefit from this potentially life-saving drug.

10.5: Mole-mass and mass-mass problems (7)

key to take with you

A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.