Lesson Explainer: Reaction Masses Chemistry

In this explainer, we will learn how to calculate the masses of reactants and products using their formula masses, molar quantities, and ratios.

Let us consider the following balanced chemical equation: 2H()+O()2HO()222ggg

The number appearing in front of each molecule in a balanced chemical equation is called the stoichiometric coefficient. Stoichiometric coefficients indicate the number of molecules or formula units of each species required so that the reaction does not violate the law of conservation of mass.

Definition: Stoichiometric Coefficients

The stoichiometric coefficient is the number appearing in front of a species in a chemical equation to indicate the number of that species present during the reaction.

When no stoichiometric coefficient appears in front of a species, the coefficient is assumed to be one. Therefore, in this reaction, two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor. However, if we were to perform this reaction, millions of hydrogen gas and oxygen gas molecules would be reacted.

Since many molecules and formula units react in any given reaction, it is much more common for chemists to recognize the stoichiometric coefficients as the amount of each substance in moles rather than individual molecules or formula units.

The relationship between the amounts in moles of two substances in a chemical reaction is called the molar ratio. Molar ratios can be written for any two species in a chemical equation, regardless of whether the species are reactants or products.

Definition: Molar Ratio

The molar ratio is the ratio of the amount in moles of one substance to the amount in moles of another substance, which are both involved in a chemical reaction, represented by the stoichiometric coefficients in a balanced chemical equation.

We can see from the balanced chemical equation for the reaction of hydrogen and oxygen to produce water that two moles of hydrogen gas react to produce two moles of water vapor. We can express this relationship, the molar ratio, in a variety of ways, as shown in the list below.

  • Molar Ratio as an Expression 22molHmolHO22 where the symbol means stoichiometrically equivalent.
  • Molar Ratio as a Ratio 2222molHmolHO22
    Or simplified to 1111molHmolHO22
  • Molar Ratio as a Conversion Factor 2211.molHmolHOorsimpliedtomolHmolHO2222 Or the inverse 2211.molHOmolHorsimpliedtomolHOmolH2222

Example 1: Deducing the Molar Ratio of Reactants from a Balanced Reaction Equation

Methane burns in oxygen according to the following equation: CH+2OCO+2HO4222

  1. What is the molar ratio of methane to oxygen?
  2. What is the molar ratio of methane to carbon dioxide?
  3. What is the molar ratio of oxygen to water?

Answer

All three questions ask us to determine the molar ratio of two substances using a balanced chemical equation. The molar ratio relates the number of moles of one substance to the number of moles of another substance involved in the same chemical reaction. The stoichiometric coefficient, the number that appears in front of a species in a chemical equation, represents the number of moles of that species necessary for a complete reaction.

Part 1

To deduce the molar ratio of methane (CH4) to oxygen (O2), we need to identify the stoichiometric coefficient in front of each species in the chemical equation. Methane has no value written in front of it. Therefore, its stoichiometric coefficient is one. The stoichiometric coefficient written in front of the oxygen is two. Hence, we write these two values as a ratio, giving us the molar ratio of methane to oxygen, 12.

Part 2

We have already determined the stoichiometric coefficient of methane to be one. Carbon dioxide (CO2) has no value written in front of it in the chemical equation. Therefore, its stoichiometric coefficient is also one. Hence, we write these two values as a ratio, giving us the molar ratio of methane to carbon dioxide, 11.

Part 3

The stoichiometric coefficients in front of the oxygen (O2) and water (HO2) are both two. We can write these values as the ratio 22. However, ratios should always be simplified. Therefore, the molar ratio of oxygen to water is 11.

We can use molar ratios written as a conversion factor to convert between the moles of two different substances. Suppose we wanted to know how many moles of water vapor could be produced from eight moles of hydrogen gas and excess oxygen gas: 2H()+O()2HO()222ggg

We can use dimensional analysis to convert the eight moles of hydrogen gas into moles of water vapor. In this process, we multiply the original value and unit by the appropriate molar ratio so that any undesired units cancel. Units that appear in both the numerator and the denominator cancel.

To perform the conversion, we multiply the moles of hydrogen by the molar ratio for hydrogen to water written as a fraction with mol H2 in the denominator: 8×11.molofHmolHOmolH222

Hence, the unit, mol H2, cancels: 8×11,molHmolHOmolH222 leaving us with the number of moles of water: 8×11=8.molHmolHOmolHmolHO2222

Notice that the molar ratio was written with moles of hydrogen in the denominator. Let us consider what would happen if the conversion factor is inverted with moles of hydrogen in the numerator: 8×11.molHmolHmolHO222

The unit mole appears in the numerator and denominator, so we might think that these units will cancel. However, it is important to consider the substance as part of the unit. Thus, mol H2 in the numerator cannot be canceled with mol HO2 in the denominator.

Example 2: Calculating the Moles of a Reactant Consumed in a Reaction Given the Moles of a Second Reactant

Acetylene (CH22) is used in welding torches. The combustion of acetylene is described by the balanced chemical equation 2CH()+5O()4CO()+2HO()22222gggg

If 8.5 moles of acetylene is burned, how many moles of oxygen gas must have been consumed in a complete reaction? Give your answer to 2 decimal places.

Answer

We need to use dimensional analysis to convert from moles of acetylene to moles of oxygen. This means that we need a conversion factor that relates moles of acetylene to moles of oxygen. This conversion factor is called the molar ratio.

The stoichiometric coefficients in a balanced chemical equation represent the number of moles of each species necessary for a complete reaction. The stoichiometric coefficients in front of CH22 and O2 are two and five respectively. The molar ratio of acetylene to oxygen, 25, can be expressed as the following conversion factors: 25,52.molCHmolOmolOmolCH222222

When performing dimensional analysis, the units that appear in both the numerator and the denominator cancel. The value that we are converting is 8.5 moles of CH22. We need the unit moles of CH22 to cancel. Thus, we should multiply the original value by the conversion factor, which has moles of CH22 in the denominator: 8.5×528.5×52.molCHmolOmolCHmolCHmolOmolCH2222222222

We then perform the calculation to determine the number of moles of oxygen gas: 8.5×52=21.25.molCHmolOmolCHmolO222222

When burned, 8.5 moles of acetylene completely reacts with 21.25 moles of oxygen gas.

Molar ratios allow converting between the amounts of two different substances in moles. However, when we wish to carry out a reaction in the laboratory, we do not measure our reactants in moles. Instead, we may use a balance to measure the masses of our reactants in grams.

Let us consider the reaction 2NaHCO()NaCO()+HO()+CO()32322ssgg

We can calculate how many grams of sodium carbonate (NaCO23) we expect to be produced if we heated 46 grams of sodium bicarbonate (NaHCO3). We know from the balanced chemical equation that 2 moles of sodium bicarbonate can produce 1 mole of sodium carbonate, as expressed by 2=1,21,12.molNaHCOmolNaCOmolNaHCOmolNaCOmolNaCOmolNaHCO323323233

The molar ratio only relates to the number of moles of each substance. We know the starting mass of sodium bicarbonate in grams, and we want to know the mass of sodium carbonate that can be produced in grams. This means that we need to convert the mass of sodium bicarbonate into moles of sodium bicarbonate so that we can use the molar ratio to compare the two compounds. Then, we need to convert the moles of sodium carbonate into grams of sodium carbonate. This series of conversions is outlined in the figure below.

We can convert between mass and moles using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

It is extremely important while working out this problem that we carefully label each substance, especially since the formulas for the two compounds in question look very similar. It is also useful to determine the molar masses of the two substances before working out the rest of the problem.

The molar mass of each compound can be calculated by summing the average molar masses of the constituent atoms:
molar mass of sodium bicarbonate: 𝑀=𝑀+𝑀+𝑀+3×𝑀𝑀=23/+1/+12/+(3×16/)𝑀=84/,()()()()()()()NaHCONaHCONaHCONaHCO333gmolgmolgmolgmolgmol
molar mass of sodium carbonate: 𝑀=2×𝑀+𝑀+3×𝑀𝑀=(2×23/)+12/+(3×16/)𝑀=106/.()()()()()()NaCONaCONaCONaCO232323gmolgmolgmolgmol

Now, we can begin solving the original problem. First, we substitute the mass and molar mass of sodium bicarbonate into the mass/mole equation: 𝑛=𝑚𝑀𝑛=4684/.()()()()NaHCONaHCONaHCONaHCO3333ggmol

Then, we determine the number of moles of sodium bicarbonate: 𝑛=4684/=0.5476.()NaHCO33ggmolmolNaHCO

Next, we can multiply the number of moles of sodium bicarbonate by the molar ratio of sodium bicarbonate to sodium carbonate: 0.5476×12.molNaHCOmolNaCOmolNaHCO3233

We wrote the molar ratio as a fraction with moles of sodium bicarbonate in the denominator so that the unit cancels: 0.5476×12.molNaHCOmolNaCOmolNaHCO3233

We are left with moles of sodium carbonate: 0.5476×12=0.2738.molNaHCOmolNaCOmolNaHCOmolNaCO323323

Finally, we can substitute the moles and molar mass of sodium carbonate into the rearranged mass/mole equation: 𝑚=𝑛𝑀𝑚=0.2738106/.()()()()NaCONaCONaCONaCO23232323molgmol

Then, we determine the mass of sodium carbonate produced: 𝑚=29.0238.()NaCO2323gofNaCO

Thus, 46 grams of sodium bicarbonate can produce just over 29 grams of sodium carbonate when heated.

Example 3: Calculating the Mass of a Reactant in a Reaction with One-to-Many Stoichiometry

Carbon tetrachloride can be synthesized by reacting chlorine with methane according to the equation CH+4ClCCl+4HCl424 [C = 12 g/mol, H = 1 g/mol, Cl = 35.5 g/mol]

If 8.0 g of HCl is produced, what mass of methane is consumed? Give your answer to the nearest 2 decimal places.

Answer

This question is asking us to convert from grams of HCl to a mass of methane (CH4). Two different substances can be related to one another via their molar ratio. The molar ratio relates the amount in moles of one substance to the amount in moles of another substance, which are both involved in a chemical reaction. The stoichiometric coefficients in a balanced chemical equation indicate the number of moles of each species involved in the reaction. When no number is written in front of a species, the coefficient is assumed to be one. Looking at the balanced chemical equation, we can determine the molar ratio of CH4 to HCl to be 1 mol CH4 to 4 mol HCl.

In order to use the molar ratio, we need to convert the mass of HCl given in the question into moles. This can be accomplished using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole. To use the equation, we first need to calculate the molar mass of HCl by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀𝑀=1/+35.5/𝑀=36.5/.()()()()()HClHClHClHClgmolgmolgmol

We can then substitute the mass and molar mass of HCl into the equation: 𝑛=8.036.5/.ggmol

Then, we can calculate the number of moles of HCl: 𝑛=8.036.5/=0.219.ggmolmolHCl

Next, we can convert the moles of HCl into moles of CH4 by multiplying by the amount of HCl in moles by the molar ratio written as a fraction. The molar ratio should be written with the moles of HCl in the denominator, 0.219×14,molHClmolCHmolHCl4 so that the unit mol HCl cancels, 0.219×14,molHClmolCHmolHCl4 leaving us with moles of CH4, 0.219×14=0.0548.molHClmolCHmolHClmolCH44

As the question is asking us to determine the mass of methane, we need to convert the amount of methane in moles into mass in grams. We can use the same mass/mole equation that we used earlier: 𝑛=𝑚𝑀, which can be rearranged to solve for the mass: 𝑚=𝑛𝑀.

We know the moles of methane, but we need to calculate the molar mass by summing the average molar masses of the constituent atoms: 𝑀=𝑀+4×𝑀𝑀=12/+(4×1/)𝑀=16/.()()()()()CHCHCHCH444gmolgmolgmol

We can then substitute the molar mass and amount of CH4 in moles into the equation: 𝑚=0.054816/.molgmol

Then, we can calculate the mass of methane: 𝑚=0.877.gofCH4

Rounding our answer to two decimal places, we determined that if 8.0 g of HCl is produced, 0.88 grams of methane must have been consumed.

We have now seen how the amount of a substance in moles can be related to the mass of the substance or the amount of another substance in moles. The amount of a substance in moles can also be related to the number of entities (atoms, molecules, units, ions, or particles) of a substance by the equation 𝑛=𝑁𝑁, where 𝑛 is the amount in moles, 𝑁 is the number of entities, and 𝑁 is Avogadro’s constant (6.022×10 mol−1). The figure below shows the relationships among the masses, amounts in moles, and entities of two substances.

Example 4: Calculating the Mass of Oxygen Required to React with a Given Number of Magnesium Atoms

How many grams of oxygen gas are necessary to react completely with 2.93×10 atoms of magnesium to yield magnesium oxide? Give your answer to 2 decimal places in scientific notation. [O = 16 g/mol, Mg = 24 g/mol]

Answer

In order to relate oxygen gas and magnesium, we need a balanced chemical equation. Magnesium and oxygen gas are the reactants and should be written on the left-hand side of the reaction arrow. Oxygen gas is a diatomic molecule when it is in its pure form and should appear as O2 in the chemical equation. Magnesium oxide, which has the chemical formula MgO, is the product and should be written on the right-hand side of the reaction arrow. Hence, the reaction is written as follows: Mg+OMgO2

As written, this chemical equation is unbalanced. We can balance it by making a list of the atoms of each element on both sides of the equation.

We can see that the magnesium atoms are balanced but the oxygen atoms are not. Therefore, we can place a coefficient of two in front of the magnesium oxide in order to balance the oxygen atoms.

Placing a coefficient of two in front of the magnesium oxide unbalances the magnesium atoms. However, they can be rebalanced by placing a coefficient of two in front of the magnesium reactant.

Stoichiometric coefficients in a balanced chemical equation indicate the number of moles of each species involved in the reaction. When no number is written in front of a species, the coefficient is assumed to be one. Looking at the above balanced chemical equation, we can see that two moles of magnesium react with one moles of oxygen. This relationship is called the molar ratio and can be used as a conversion factor to convert between the two substances.

However, the question gave us the amount of magnesium in atoms instead of moles. Therefore, we can convert a number of entities, in this case atoms, into moles using the equation 𝑛=𝑁𝑁, where 𝑛 is the amount in moles, 𝑁 is the number of entities, and 𝑁 is Avogadro’s constant (6.022×10 mol−1). We can then substitute the number of atoms and Avogadro’s constant into the equation, 𝑛=2.93×106.022×10,mol and determine the number of moles of magnesium: 𝑛=0.00487.molMg

Next, we can convert the moles of magnesium into moles of oxygen gas by multiplying by the molar ratio written as a fraction. The molar ratio should be written with the moles of magnesium in the denominator, 0.00487×12,molMgmolOmolMg2 so that the unit mol Mg cancels, 0.00487×12,molMgmolOmolMg2 leaving us with moles of O2, 0.00487×12=0.00243.molMgmolOmolMgmolO22

As the question is asking us to determine the mass of oxygen gas in grams, we need to convert the amount of oxygen gas in moles into mass in grams. To perform this conversion, we can use the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole. We then rearrange to solve for the mass: 𝑚=𝑛𝑀.

Next, we substitute with the amount and molar mass of oxygen gas (32 g/mol) into the equation, 𝑚=0.00243×32/,molgmol to determine the mass of oxygen gas, 𝑚=0.077848.gO2

The question asks us to give our answer in scientific notation rounded to two decimal places. Writing our answer appropriately, 7.78×10 grams of oxygen gas is necessary to react completely with 2.93×10 atoms of magnesium to yield magnesium oxide.

Key Points

  • In a balanced chemical equation, the stoichiometric coefficient represents the number of moles of each substance.
  • The relationship between the amounts in moles of two substances in a chemical reaction is called the molar ratio.
  • Moles of one substance can be converted into moles of another substance by multiplying by the molar ratio written as a fraction.

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