# Why can't I conserve mass instead of moles and apply ratio in this problem?

I was solving this problem:

$$S$$ combines with $$O_2$$ to form $$SO_2$$ and $$SO_3$$. If $$10g$$ of $$S$$ is mixed with $$12g$$ of $$O_2$$, what mass of $$SO_2$$ will be formed, so that neither $$O$$ nor $$S$$ will be left at the end of reaction?

My understanding was that in normal reaction where there would have been sufficient amount of reactants, sufficient amount of products would have been formed according the the ratios of their masses and the total mass. So I used the below methodology:

Total mass of reactant = Total mass of product = 22g

GMM($$SO_2$$) = 64

GMM($$SO_3$$) = 80

Mass of $$SO_2$$ produced = $$\frac{64}{64+80} * 22$$

= $$0.4444*22$$ = $$9.7$$$$10$$

But this is equal to the mass of $$SO_3$$ produced and thus is the opposite of the answer which is obtained upon conserving moles.

The other answers have explained how to solve this problem correctly (using either moles or mass ratios), but not what you did wrong. So let's take a closer look at your reasoning (with my annotation inserted in red):

My understanding was that in normal reaction where there would have been sufficient amount of reactants, sufficient amount of products would have been formed according the the ratios of their masses and the total mass. So I used the below methodology:

Total mass of reactant = Total mass of product = 22g

GMM($$SO_2$$) = 64

GMM($$SO_3$$) = 80

Mass of $$SO_2$$ produced = $$\frac{64}{64+80} * 22$$ $$\quad \color{red}{\leftarrow \text{THE PROBLEM IS HERE}}$$

= $$0.4444*22$$ = $$9.7$$$$10$$

On the line I've marked with the note above, you've assume that the fraction of the total product mass that will be $$\ce{SO2}$$ equals the mass of one $$\ce{SO2}$$ molecule divided by the summed masses of one $$\ce{SO2}$$ molecule plus one $$\ce{SO3}$$ molecule.

In other words, you're assuming that exactly 50% of the molecules produced in the reaction will be $$\ce{SO2}$$, while the other 50% will be $$\ce{SO3}$$.

Note that the input masses of $$\ce{S}$$ and $$\ce{O}$$ don't enter your calculations at all! That's a sure sign that something in your reasoning is wrong, since it means that changing the ratio of the reactants doesn't change the ratio of the products that you've calculated, even though in reality it very much should.

In other words, the answer you got is nonsense. The fact that it happens to match something actually related to the correct answer is just a pure coincidence, no doubt brought about by the strong law of small numbers and the exercise being deliberately designed around simple integer ratios to make it easy to calculate.

In particular, the exercise says that "neither $$\ce{O}$$ nor $$\ce{S}$$ will be left at the end of reaction." This means that all the reactants are (assumed to be) consumed in the reaction, and thus that the ratio of the reactants provided definitely must affect the ratio of the products. Specifically, each $$\ce{S}$$ atom must go into either $$\ce{SO2}$$ or $$\ce{SO3}$$ and your task is figuring out the ratio in which these are produced so that each $$\ce{O}$$ atom also goes into one of these two products.

This is pretty easy to do if you convert everything into moles first, but it can be done using just masses, too. One easy way, as others (such as Maurice) have already noted, is to first assume that all of the sulfur goes into $$\ce{SO2}$$, consuming as much of the supplied oxygen as it needs to do so.

Let's do that, writing out all the steps and using slightly more accurate molecular masses just for fun:

Molecular mass of sulfur = $$M_{\ce{S}}$$$$\ce{32.065 u}$$

Molecular mass of oxygen = $$M_{\ce{O}}$$$$\ce{15.999 u}$$

Mass fraction of sulfur in $$\ce{SO2}$$ = $$x$$ = $$\dfrac{M_{\ce{S}}}{M_{\ce{S}} + 2M_{\ce{O}}}$$$$\dfrac{\ce{32.065 u}}{\ce{32.065 u} + 2 \cdot \ce{15.999 u}}$$$$0.50052$$

Mass of $$\ce{SO2}$$ produced by reacting $$\ce{10 g}$$ of sulfur with sufficient oxygen = $$m'_{\ce{SO2}}$$ = $$\ce{10 g} \mathbin/ x$$$$\ce{19.979 g}$$

Mass of oxygen consumed in this reaction = $$m'_{\ce{O}}$$ = $$(1 - x) \cdot m'_{\ce{SO2}}$$$$0.49948 \cdot \ce{19.979 g}$$$$\ce{9.9791 g}$$

(If you had used $$M_{\ce{S}} = \ce{32 u}$$ and $$M_{\ce{O}} = \ce{16 u}$$, this would've of course worked out to $$x = \frac12$$ exactly, and thus to $$m'_{\ce{O}} = \ce{10 g}$$, same as the input mass of sulfur. By using more accurate atomic masses we avoided this collision of small numbers.)

Now we just need to figure out how much oxygen is left and how much of the $$\ce{SO2}$$ needs to be turned into $$\ce{SO3}$$ in order to consume all of it:

Mass of remaining oxygen after reacting all sulfur into $$\ce{SO2}$$ = $$m''_{\ce{O}}$$ = $$\ce{12 g} - m'_{\ce{O}}$$$$\ce{2.0209 g}$$

Mass fraction of a single oxygen atom in $$\ce{SO3}$$ = $$y$$ = $$\dfrac{M_{\ce{O}}}{M_{\ce{S}} + 3M_{\ce{O}}}$$$$\dfrac{\ce{15.999 u}}{\ce{32.065 u} + 3 \cdot \ce{15.999 u}}$$$$0.19983$$

Final mass of $$\ce{SO3}$$ produced by reacting all remaining oxygen with $$\ce{SO2}$$ to form $$\ce{SO3}$$ = $$m_{\ce{SO3}}$$ = $$m''_{\ce{O}} \mathbin/ y$$$$\ce{10.113 g}$$

(Again, with $$M_{\ce{S}} = \ce{32 u}$$ and $$M_{\ce{O}} = \ce{16 u}$$ this would've worked out to $$m''_{\ce{O}} = \ce{2 g}$$ and $$y = \frac15$$ and thus $$m_{\ce{SO3}} = \ce{10 g}$$ exactly.)

Mass of $$\ce{SO2}$$ converted into $$\ce{SO3}$$ by reacting with the remaining oxygen = $$m''_{\ce{SO2}} = (1 - y) \cdot m_{\ce{SO3}}$$$$0.80017 \cdot \ce{10.113 g}$$$$\ce{8.0921 g}$$

Final mass of $$\ce{SO2}$$ remaining after all sulfur and all oxygen has been consumed = $$m_{\ce{SO2}}$$ = $$m'_{\ce{SO2}} - m''_{\ce{SO2}}$$$$\ce{11.887 g}$$

Of course we can (and should) also do a final consistency check to make sure that the total mass of the products actually equals the total mass of the reactants: $$m_{\ce{SO2}} + m_{\ce{SO3}}$$$$\ce{11.887 g} + \ce{10.113 g}$$$$\ce{22.000 g}$$, just as it should be.

Note that, since I performed multiple rounding steps during this calculation, the fact that the total mass happened to round to exactly $$\ce{22.000 g}$$ is mostly luck. I could've just as easily gotten an answer of $$\ce{21.999 g}$$ or $$\ce{22.001 g}$$ or maybe even further off. But as long as the rounding error stays in the last digit, you can be pretty confident that the calculation is right. Or at least not wrong in such a way that it would violate the conservation of mass.

Of course you could also do the calculation with a few more significant digits in the intermediate results and only round to the input precision — which in this case, if we assume the input masses to be exact, is limited by the precision of $$M_{\ce{S}}$$ and $$M_{\ce{O}}$$ — at the end.

• Thank you very much for the detailed answer. I just had one more doubt.Can you please confirm if I am correct in thinking that even if there was one mole of $SO_2$ and one mole of $SO_3$ produced in a balanced reaction my method would not work as the reaction proceeds in two steps? Commented Jun 27 at 17:17
• This is nonsensical. the numbers were chosen to make the math easy; with 2 sig. figs. the use of super-accurate MW is not appropriate. To the above comment: This reaction is a two-step process. The math question is whether this is a unique answer regardless of the process. you must determine if your method works regardless. Commented Jun 27 at 20:17

This problem can be solved by figuring out the mass ratios. 32 grams of sulfur react with 32 grams of oxygen to form sulfur dioxide (1:1 mass ratio). 64 grams of sulfur dioxide react with 16 grams of oxygen to form sulfur trioxide (4:1 mass ratio).

So to turn all the sulfur (10 grams) into sulfur dioxide, you need 10 grams of oxygen (and this yields 20 grams, because 10 + 10 = 20). The remaining 2 grams of oxygen will react with 8 grams of sulfur dioxide, leaving 12 grams of the latter at the end.

Contemplating this problem you might think you could skip the mole concept, but the rest of the world of chemistry would convince you otherwise.

• U did not skip mole concept. S and O2 luckily have the same MW and you knew that or looked it up. Commented Jun 27 at 20:08
• @jimchmst True. But historically, mass relationships were known before the atomic age.
– Karsten
Commented Jun 27 at 20:34

Reactions of interest:

$$\ce{S(s) +O2(g)->SO2(g)}\;\;\;\;\;\;\;\;\;\;(1)$$

$$\ce{SO2(g) +1/2 O2(g)->SO3(g)}\;\;\;(2)$$

Arbitrary symbolism:

A = $$\ce{S(s)}$$

B = $$\ce{O2(g)}$$

C = $$\ce{SO2(g)}$$

D = $$\ce{SO3(g)}$$

Calculation of initial amounts:

$$n_{\mathrm{A_O}}=\frac{m_{\mathrm{A_O}}}{M_A}=\frac{\pu{10g}}{\pu{32g/mol}}=\pu{0.3125mol}$$

$$n_{\mathrm{B_O}}=\frac{m_{\mathrm{B_O}}}{M_\mathrm{B}}=\frac{\pu{12g}}{\pu{32g/mol}}=\pu{0.3750mol}$$

$$n_{\mathrm{C_O}}=0$$

$$n_{\mathrm{D_O}}=0$$

Calculation of final amounts in reaction (1):

$$n_{\mathrm{A_1}}=n_{\mathrm{A_O}}(1-X_\mathrm{A})=0$$

$$n_{\mathrm{B_1}}=n_{\mathrm{A_O}}(\Theta_{\mathrm{B_O}}-X_\mathrm{A})=\pu{0.3125mol}\left(\frac{\pu{0.3750mol}}{\pu{0.3125mol}}-1\right)=\pu{0.06250mol}$$

$$n_{\mathrm{C_1}}=n_{\mathrm{A_O}}(\Theta_{\mathrm{C_O}}+X_\mathrm{A})=\pu{0.3125mol}$$

$$n_{\mathrm{D_1}}=0$$

Calculation of final amounts in reaction (2):

$$n_{\mathrm{A_2}}=0$$

$$n_{\mathrm{B_2}}=n_{\mathrm{B_1}}\left(1-X_\mathrm{B}\right)=0$$

$$n_{\mathrm{C_2}}=n_{\mathrm{B_1}}\left(\Theta_{\mathrm{C_1}}-2X_\mathrm{B}\right)=\pu{0.06250mol}\left(\frac{\pu{0.3125mol}}{\pu{0.06250mol}}-2\right)=\pu{0.1875mol}$$

$$n_{\mathrm{D_2}}=n_{\mathrm{B_1}}\left(\Theta_{\mathrm{D_1}}+2X_\mathrm{B}\right)=(0.06250)(2)\pu{mol}=\pu{0.1250mol}$$

Final mass of each product:

$$m_{\mathrm{C_2}}=n_{\mathrm{C_2}}\;M_\mathrm{C}=(\pu{0.1875mol})(\pu{64g/mol})=\pu{12g}$$

$$m_{\mathrm{D_2}}=n_{\mathrm{D_2}}\;M_\mathrm{D}=(\pu{0.1250mol})(\pu{80g/mol})=\pu{10g}$$

Total final mass of products:

$$m_2=m_{\mathrm{C_2}}+m_{\mathrm{D_2}}=(12+10)\pu{g}=\pu{22g}$$

Mass conservation verification:

$$m_0=m_{\mathrm{A_O}}+m_{\mathrm{B_O}}=\pu{22g}=m_{\mathrm{C_2}}+m_{\mathrm{D_2}}=m_2$$

Here is a quicker solution, with the same reactions ($$1$$) and ($$2$$) as defined by Sam$$202$$.

$$\ce{S(s) + O2(g)⟶SO2(g)} \tag{1}$$

$$\ce{SO2(g) + 1/2 O2(g) ⟶ SO3(g)} \tag{2}$$

$$10$$ g sulfur is reacted with $$\ce{O2}$$ via ($$1$$). In mole, this is $$10/32 = 0.3125$$ mol $$\ce{S}$$. This $$0.3125$$ mol $$\ce{S}$$ reacts in ($$1$$} with the same amount of $$\ce{O2}$$($$0.3125$$ mol) and produces the same amount $$0.3125$$ mol $$\ce{SO2}$$, or in mass $$0.3125 · 64$$ g = $$20$$ g $$\ce{SO2}$$.

Let's study the amount of $$\ce{O2}$$ consumed in ($$1$$). $$0.3125$$ mol $$\ce{S}$$ reacts with $$0.3125$$ mol $$\ce{O2}$$, or in mass : $$0.3125 · 32 g = 10$$ g $$\ce{O2}$$. This $$10$$ g $$\ce{O2}$$ is less than the mass of $$\ce{O2}$$ available ($$12$$ g $$\ce{O2}$$). It means that $$12 - 10 = 2$$ g $$\ce{O2}$$ reacts with a fraction of the $$\ce{SO2}$$ synthesized, to produce $$\ce{SO3}$$ according to ($$2$$). How big is this fraction of $$\ce{SO2}$$ ?

Well ! $$2$$ g $$\ce{O2}$$ is also $$2/32$$ mole = $$0.0625$$ mol $$\ce{O2}$$. In ($$2$$), this amount $$0.0625$$ mol $$\ce{O2}$$ reacts with the double, or $$0.125$$ mol $$\ce{SO2}$$. The remaining $$\ce{SO2}$$ having not reacted in ($$2$$) is : $$0.3125$$ mol - $$0.125$$ mol = $$0.1875$$ mol $$\ce{SO2}$$, which weighs $$0.1875 · 64$$ g = $$12$$ g $$\ce{SO2}$$ : first final result.

It is not required, but interesting to calculate the mass of $$\ce{SO3}$$ produced. We know that $$2$$ g $$\ce{O2}$$ will be used to produce $$\ce{SO3}$$ via ($$2$$). $$2$$ g, or $$0.0625$$ mol $$\ce{O2}$$, reacts with the double, or $$0.125$$ mol $$\ce{SO2}$$. And it produces $$0.125$$ mole $$\ce{SO3}$$. This weighs $$0.125 · 80$$ g = $$10$$ g $$\ce{SO3}$$ : Second final result.

Something surprising : the final weights ($$12$$ g and $$10$$ g) are the same as the original weights ($$10$$ g and $$12$$ g).

This is really a logic and math problem because the chemistry is a bit specious, but it helps to know chemistry. Sulfur combines with O2 to give SO2 and possibly SO not a mix of SO2 and SO3. The oxidation to SO3 requires a catalyst [contact process] or a different mechanism involving nitrogen oxides or possibly ozone.

First oxidize all the S to SO2; 10/32S +10/32O2 = 10/32SO2 leaving 2/32O2. Now the catalyst is introduced and the remaining O2 oxidizes SO2 to SO3.

2/32O2 [4/32 O] + 4/32SO2 = 4/32SO3 Final mix = 6/32SO2 [12g] + 4/32SO3 [10g]; total =22g. It is assumed the reactions proceed to completion but they are probably run with excess O2 because it is available.

This is a nice clean answer; actual situations such as the processes in the cylinder of an internal combustion engine are much more involved.

The problem is with this:

Mass of $$SO_2$$ produced = $$\frac{64}{64+80} * 22$$

That is an invalid assumption, it only works if you produce the same amount of $$SO_2$$ and $$SO_3$$. If you use variables x and y to represent amounts of $$SO_2$$ and $$SO_3$$ respectively you could use this formula:

$$\frac{64x}{64x+80y} * 22$$

Finding 'x' and 'y' is the point of the exercise. Your assumption is that x = y which is not the case.

If you want to think of it in a different way, the molecular weight of Sulfur is very close to twice the molecular weight of Oxygen (32.065 vs. 15.999). The starting ratio of atoms of S and O would be 10 Sulfur to 24 oxygen. If you use all the Sulfur to make $$SO_2$$ that would use 20 of the oxygen. That leaves you with 4 oxygen to combine with the $$SO_2$$ to form $$SO_3$$ leaving you with 6 $$SO_2$$ and 4 $$SO_3$$. Put those number in place of 'x' and 'y' in the equation above:

$$\frac{64*6}{64 * 6 + 80 * 4} * 22$$

Or $$\frac{384}{384+320} * 22$$

Or $$\frac{384}{704} * 22$$

Or $$\frac{6}{11} * 22$$

Or $$12$$

If the problem stated that you started with 14g of $$O_2$$ that would alter the calculation. Now you would start with 10 Sulfur and 28 Oxygen, combining 20 Oxygen with 10 Sulfur gives you 10 $$SO_2$$ and leaves you with 8 Oxygen. Combining 8 oxygen with 8 $$SO_2$$ gives you 8 $$SO_3$$ with 2 $$SO_2$$ remaining. That wouldn't change your original formula, but now x is 2 and y is 8:

$$\frac{64*2}{64 * 2 + 80 * 8} * 22$$

Which is $$\frac{128}{768} * 22$$ or about to $$3.666$$ grams of $$SO_2$$.

As a follow up answer, it is possible to calculate the correct final product mass fractions ($$y_{\mathrm{C_2}}, y_{\mathrm{D_2}}$$) and then final product mass ($$m_{\mathrm{C_2}}, m_{\mathrm{D_2}}$$) using the following expressions that are only in terms of $$\;$$ molar mass ($$M_i$$), initial mass ($$m_{i_O}$$), initial mass fractions ($$y_{i_O}$$), and total initial mass $$(m_o)$$, thus skipping use of moles:

Arbitrary symbolism:

A = $$\ce{S(s)}$$

B = $$\ce{O2(g)}$$

C = $$\ce{SO2(g)}$$

D = $$\ce{SO3(g)}$$

Known values:

$$M_\mathrm{A}=\pu{32g/mol}$$

$$M_\mathrm{C}=\pu{64g/mol}$$

$$M_\mathrm{D}=\pu{80g/mol}$$

$$m_{\mathrm{A_O}}=\pu{10g}$$

$$m_{\mathrm{B_O}}=\pu{12g}$$

Calculating initial mass fraction of A:

$$y_{\mathrm{A_O}}=\frac{m_{\mathrm{A_o}}}{m_{\mathrm{A_o}}+\;m_{\mathrm{B_o}}}=\frac{\pu{10g}}{\pu{22g}}=\frac{5}{11}$$

Calculating final mass fractions of C and D:

$$y_{\mathrm{C_2}}=y_{\mathrm{A_O}}\left(\frac{\mathrm{M_C}}{\mathrm{M_A}}\right)\left(3-2\;\frac{m_{\mathrm{B_o}}}{m_{\mathrm{A_o}}}\right)=\frac{5}{11}\left(\frac{\pu{64g/mol}}{\pu{32g/mol}}\right)\left(3-2\;\frac{\pu{12g}}{\pu{10g}}\right)=\color{red}{\pmb{\frac{6}{11}}}$$

$$y_{\mathrm{D_2}}=2y_{\mathrm{A_O}}\left(\frac{\mathrm{M_D}}{\mathrm{M_A}}\right)\left(\frac{m_{\mathrm{B_o}}}{m_{\mathrm{A_o}}}-1\right)=2\left(\frac{5}{11}\right)\left(\frac{\pu{80g/mol}}{\pu{32g/mol}}\right)\left(\frac{\pu{12g}}{\pu{10g}}-1\right)=\color{red}{\pmb{\frac{5}{11}}}$$

Calculating final mass of C and D:

$$m_{\mathrm{C_2}}=y_{\mathrm{C_2}}\;m_o=\left(\frac{6}{11}\right)\left(\pu{22g}\right)=\color{red}{\pmb{\pu{12g}}}$$

$$m_{\mathrm{D_2}}=y_{\mathrm{D_2}}\;m_o=\left(\frac{5}{11}\right)\left(\pu{22g}\right)=\color{red}{\pmb{\pu{10g}}}$$