# What are practical examples of simultaneous measuring of quantities?

I am a maths teacher and I'm looking for practical examples of measuring quantities simultaneously, so that mathematically one needs to construct and solve a system of equations. I thought that in the world of chemistry there could well be many possible such problems, but alas I have no experience of chemistry, hence I thought to ask here.

I searched online and found two possible situations.

1. Given an alloy of say two metals, we can use the combined weight as one equation and the combined volume as a second equation and hence we can find the amount of each metal in the mixture. What I like about this is the simplicity whereas I assume it is much more difficult to extract the metals separately.

2. Given a mixture of two gases, we can see how much light of one frequency is absorbed and again for a second frequency which allows the amount of both gases to be simultaneously found.

What other such situations are there is chemistry; when would one actually need to do such simultaneous calculations?

• Careful: The volume of an alloy is not just the sum of the volumes of the constituents. – Karl Apr 30 '20 at 19:08
• You could use an emulsion as an example. Or a foam. – Karl Apr 30 '20 at 19:16
• For the math, do the students know just algebra or do they know some calculus too? // gas/light notion - as light goes through gas the intensity decreases. Say 1 inch of travel cuts light down by 50%, 2 inches yields 25% and so on. One would really need to understand the basics of calculus to understand how the intensity/concentration relationship works. – MaxW Apr 30 '20 at 19:50
• I'd assume that the students have had no chemistry and that the problems should be formulated in a way that presumes no knowledge of chemistry? – MaxW Apr 30 '20 at 19:58
• @Max That's right. They won't know calculus at that point, basic algebra. My idea was to give a little statement of where the equations come from and then they need to solve them. – Geoff Apr 30 '20 at 20:12

You may also give the weight of a sample made of an alloy $$\ce{Zn-Mg}$$. Then you dip it into some diluted $$\ce{HCl}$$. Both metals will react simultaneously according to $$\ce{Zn + 2 HCl -> H_2 + ZnCl_2}$$ and $$\ce{Mg + 2 HCl -> H_2 + MgCl_2}$$ This will produce an important amount of $$\ce{H_2}$$ gas. You can measure its volume. So you have two measured data, the original mass and the volume of gas. It is enough for calculating the proportion of $$\ce{Zn}$$ and $$\ce{Mg}$$ in the original sample.

Example : Take a mixture of $$0.1$$ mol $$\ce{Zn}$$ + $$0.3$$ mol $$\ce{Mg}$$. Of course the molar masses of $$\ce{Zn}$$ and $$\ce{Mg}$$ are respectively $$65.39$$ g/mol and $$24.32$$ g/mol. The total mass is : $$6.54$$ g + $$7.29$$ g = $$13.83$$ g. Suppose the temperature and the pressure of the gas is such that $$1$$ mole gas occupies $$24$$ L. So $$0.1$$ mol $$\ce{Zn}$$ will produce $$0.1$$ mol $$\ce{H_2}$$ which occupies $$2.4$$ L gas. Then $$0.3$$ mol $$\ce{Mg}$$ will produce $$0.3$$ mol $$\ce{H_2}$$, or $$7.2$$ L gas. The total volume of $$\ce{H_2}$$ is $$2.4$$ L + $$7.2$$ L = $$9.6$$ L.

If now you simply give the following data to your student : Total mass = $$13.83$$ g, and total volume of $$\ce{H_2}$$ = $$9.6$$ L, they should discover two equations with two unknowns. Of course you must tell them that $$1$$ mole gas occupies $$24$$ L. If x is the number of mole of Zn, and y the number of moles of Mg, these equations are : $$\ce{x·65.39 + y·24.32 = 13.83}$$ $$\ce{x + y = 9.6/2.4}$$ When these equations are solved, the student still has to calculate the mass of the two metals.

You may of course change the choice of the unknowns, and the nature of the metals.

• I think that if you state in the problem that 65.4 grams of pure zinc gives 24 liters of gas and that 24.3 grams of pure magnesium gives 24 liters of gas then you untangle the chemistry and reduce the problem to a math problem. – MaxW Apr 30 '20 at 22:02
• @MaxW. Yes. I have reduced the problem to a math problem. I did it on purpose, because Geoff required a math problem, not a chemistry problem ! – Maurice May 1 '20 at 10:00
• I was thinking let x= grams Cu, y = grams Mg. Cu-> 3.70 L H2/(g Cu), Mg->0.988 L H2/(g Mg) $$x + y = 13.833$$ $$x \times 3.70 + y\times0.988 = 9.6$$ That avoids any discussion of moles. – MaxW May 1 '20 at 23:06
• @MaxW. Well ! You should know that Copper does not react with Copper !! And does not produce any H2. – Maurice May 2 '20 at 12:04
• Dah... I don't know how I got Cu in my mind. Thanks for the correction. The numbers in my comment above are for Zn. – MaxW May 2 '20 at 16:20

In chemistry the Principle of Atom Conservation says that there must be the same number of like atoms on both sides of the equation. Balance the following chemical reaction by determining the appropriate values for $$a, b, c, d$$ and $$e$$ using Diophantine linear equations:

$$\ce{aS + bHNO3 <=> cH2SO4 + dNO2 + eH2O}$$ $$\text{i.e. Sulfur + Nitric acid}\ce{<=>} \text{Sulfuric acid + Nitrogen dioxide + water}$$

• For example, the same number of like atoms on both sides of the equation would mean 6 oxygen atoms on the left and 6 oxygen atoms on the right.

• The subscripts on the atoms indicate the number of that kind of atom in the molecule. Thus $$\ce{H2SO4}$$ contains 2 Hydrogen atoms(H), 1 sulfur atom(S), and 4 oxygen atoms(O).

• Standard form dictates that the greatest common multiple of the coefficients must be 1.

HINT For hydrogen(H): $$b = 2*c +2*e$$

A nastier equation would be:

$$\ce{aK2Cr2O7 + bH2SO4 + cSO2 -> dCr2(SO4)3 + eK2SO4 + fH2O}$$

• If you can put three different answers to the same question, then you should write any. – Mithoron May 1 '20 at 21:36
• @Mithoron - I changed all the answers to community wiki so I wouldn't be unethically be getting 2 extra reputation points. Happy? // I only separated the problems so that any discussion about a problem type would be on that problem. I wasn't trying to do anything nefarious. – MaxW May 1 '20 at 22:29
• My point is rather that it's obviously too broad. In such cases sometimes there's equally broad answer given, but adding multiple examples just isn't doing it. – Mithoron May 1 '20 at 23:10
• @Mithoron - I found the question intriguing. - Can you write some chemistry problems that use simultaneous equations that can be solved with no real understanding of chemistry? Maurice's Cu/Mg problem is really nice if you fiddle with it just a bit. – MaxW May 1 '20 at 23:16
• @MaxW. I never proposed a problem with Cu and Mg. Copper will not react with HCl. No ! I proposed a mixture Mg-Zn. These two metals do react with HCl. – Maurice May 2 '20 at 12:08

Acetic acid, $$\ce{CH3COOH}$$, ionizes in water according to the following chemical equation to produce a hydronium ion, $$\ce{H3O+}$$, and an acetate anion, $$\ce{CH3COO-}$$.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

The mathematical equation for the equilibrium is:

$$1.8\cdot10^{-5} =\dfrac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$

However water itself also ionizes according to the chemical equation:

$$\ce{ 2H2O <=> H3O+ + OH-}$$

The mathematical equation for that equilibrium is:

$$\ce{1\cdot10^{-14} = [H3O+][OH-]}$$

where the brackets, [], indicate concentration.

If acetic acid is initially added at a concentration of 0.1, and $$\ce{[H+] = [OH-] = 1\cdot10^{-7}}$$ what is the concentration of $$\ce{H3O+}$$ when the system reaches equilbrium?

HINTS:
(1) At equilibrium $$0.1 = \ce{[CH3COOH] + [CH3COO-]}$$ (2) Charges must balance in the solution so: $$\ce{[H3O+] = [CH3COO-] + [OH-]}$$

• I offered the hints and use a nebulous term "concentration" to try to take the required chemistry knowledge out of the problem. // The solution to this will require solving a quadratic equation. – MaxW May 1 '20 at 0:29

I think a useful problem with the properties you describe is finding how a acid with several ionized forms changes with pH in the medium. Recently I explained how to do it for $$\ce{CO2}$$ here. Look How to calculate bicarbonate and carbonate from total alkalinity. The general solution scales nicely with complexity of the system:

ACID TYPE       DIFFERENT FORMS                 GENERAL SOLUTION                                                                EXAMPLE
Monoprotic      2 (undissociated, +1)           Fraction, degree 1 polynomial at denominator, one of its terms at numerator     Acetic acid
Diprotic        3 (undissociated, +1, +2)       Fraction, degree 2 polynomial at denominator, one of its terms at numerator     Carbonic Acid
Triprotic       4 (undissociated, +1, +2, +3)   Fraction, degree 3 polynomial at denominator, one of its terms at numerator     Phosphoric acid
...             ...                             ...                                                                             ...


Plenty of aplications too. For example, explains why ocean acidification affects calcifying organisms.

Copper, $$\ce{Cu}$$, is oxidized by concentrated nitric acid, $$\ce{HNO3}$$, to produce $$\ce{Cu^{2+}}$$ ions. The nitric acid is reduced to nitrogen dioxide, $$\ce{NO2}$$, or nitric oxide, $$\ce{NO}$$, according to the following chemical equations.

$$\ce{Cu + 4HNO3 -> Cu(NO3)2 + 2NO2(gas) + 2H2O}\tag{1}$$

i.e. When 1 atom of copper is dissolved then 2 molecules of $$\ce{NO2}$$ are formed.

$$\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO(gas) + 4H2O}\tag{2}$$

i.e. When 3 atoms of copper are dissolved then 2 molecules of $$\ce{NO}$$ are formed.

When the reaction was complete by volume 63% of the gas was $$\ce{NO2}$$ and 37% of the gas was $$\ce{NO}$$, thus the ratio $$\ce{NO2:NO :: 63:37}$$. What percentage of copper dissolved by the first reaction?

• Is the answer here 36.2% ? – Geoff May 1 '20 at 9:08
• @Geoff- yes 36.2 – MaxW May 1 '20 at 9:56