(This is a reprint from the September 2007 issue of Chem 13 News, pages 4-5.)

*When a 10.0-g sample of a mixture of CH _{4} and C_{2}H_{6} is burned in excess oxygen, exactly 525 kJ of heat is produced. What is the percentage by mass of CH_{4} in the original mixture?*

The molar mass for CH_{4} is 16.042 g mol^{-1}; and the molar mass for C_{2}H_{6} is 30.068 g mol^{-1}.

∆H = −890.4 kJ (per mol CH_{4})

∆H = −1560.0 kJ (per mol C_{2}H_{6})

Only 17% of students answered this question correctly on the 2007 CHEM 13 NEWS Exam. (Interestingly, 17% is also the answer to the question!!)

It has been our experience that many students struggle with stoichiometry problems involving mixtures, particularly when two components in the mixture contribute to the amount of product (or heat) obtained. A common error made by many students when solving a problem of this type is that they add the two chemical equations together to obtain a single chemical equation involving both components of the mixture. If the two chemical equations given above are added together, we obtain the equation below.

∆H = −2450.4

While the chemical equation above is balanced, it is wrong! It implies that both CH_{4} and C_{2}H_{6} must be present, and furthermore that they must be present in a 1:1 ratio in order to produce CO_{2} and H_{2}O. It should be perfectly clear from the chemical equations that were given that CH_{4} will react with O_{2} to give CO_{2} and H_{2}O, whether or not any C_{2}H_{6} is present. (Similarly, C_{2}H_{6} will react with O_{2} whether or not any CH_{4} is present.) The two combustion reactions are independent reactions and must not be combined, no matter how tempting it might be to add them.

Perhaps many students do not fully appreciate the difference between simultaneous reactions and consecutive reactions. Consecutive reactions are reactions that occur sequentially to yield products, whereas simultaneous reactions are a set of reactions that occur simultaneously and independently of each other. An example of a set of consecutive reactions is given below for the step-wise oxidation of HBr:

HBr + O_{2} → HOOBr (1)

HOOBr + HBr → 2 HOBr (2)

HOBr + HBr → H_{2}O + Br_{2} (3)

Evidence that the three reactions (1)-(3) form a sequence is the fact that a product from an earlier reaction is a reactant in a subsequent reaction. For example, reaction (2) cannot occur until HOOBr is produced by reaction (1), and similarly, reaction (3) cannot occur until some HOBr is produced by reaction (2). The chemical equations for a set of consecutive reactions can be added algebraically to obtain a chemical equation for the “net” or “overall” process. For the set of reactions given above, the correct combination is (1) + (2) + 2×(3), which gives the result below.

4 HBr + O_{2} → 2 H_{2}O + 2 Br_{2} (4)

Note that equation (3) must be doubled before the equations are added together. Although the overall chemical equation (4) “hides” some of the chemistry involved, it correctly shows that HBr and O_{2} are consumed in the ratio 4:1 and that H_{2}O and Br_{2} are produced in a 2:2, or 1:1, ratio.

The combustion reactions for CH_{4} and C_{2}H_{6} are not consecutive reactions; they are simultaneous reactions. Simultaneous reactions must never be added together. The amount of product (or heat) obtained from each reaction is calculated independently, and the total amount of product (or heat) is calculated by adding the two amounts together at the end, as shown below.

**Approach #1: **

We are asked to determine the amount of CH_{4} in the sample, so let “x” be the mass of CH_{4}. The mass of C_{2}H_{6} must be 10.00 − x. Therefore, the number of moles of CH_{4} is (x/16.042) and the number of moles of C_{2}H_{6} is (10.00 − x)/30.068.

The combustion reactions tell us that we get 890.4 kJ of heat for every mole of CH_{4} and 1560.0 kJ of heat for every mole of C_{2}H_{6}. If q_{1} and q_{2} represent the amounts of heat we get from (x/16.042) moles of CH_{4} and (10.00 − x)/30.068 moles of C_{2}H_{6}, respectively, then we have:

The total amount of heat is q_{1} + q_{2} = 525 kJ, and so:

Solving for x, we obtain x = 1.72 and mass percentage of CH_{4} in the original mixture is approximately 17%.

**Approach #2:**

We first calculate the amounts of heat that would be obtained if the sample were pure CH_{4} or pure C_{2}H_{6}:

If “x” represents the fraction of the total mass that is CH_{4}, then q_{1} = 555.0x and q_{2} = 518.8(1−x). Therefore:

525 = 555.0 x + 518.8 (1 − x)

x = 0.17

Therefore, the sample is 17% CH_{4} by mass.

Actually, these two approaches are only superficially different. Both approaches consider how much heat is obtained from each reaction and those two quantities are then added together and set equal to the total amount of heat. The difference between the two approaches is only in what “x” represents. Another approach is to let “x” and “y” be the number of moles of CH_{4} and C_{2}H_{6}, respectively. The mass of the sample is 16.042x + 30.068y = 10.00 and the quantity of heat released is 890.4x + 1560.0y = 525. In this approach, one must solve two equations in two unknowns.

Consider applying one of these approaches to the following problem, which is slightly more complicated than problem #35 on the 2007 CHEM 13 NEWS Exam.

*A 2.00-g sample of a mixture of CaCl _{2} and RbCl is treated with excess AgNO_{3}(aq), causing AgCl(s) to precipitate from the solution. If the mass of AgCl obtained is 3.45 g, then what is the percentage by mass of RbCl in the original mixture? (Answer: 61.4% RbCl)*