Chemistry Corner

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This is a "fabulous four step" stoichiometry problem where, given the amounts of one or more substances in the reaction, you are asked to calculate the amount of another substance.

Step 1: Balanced reaction

Step 2: Moles of "given" substance(s)

Step 3. Moles of "desired" substance such as a product

Step 4: Convert Step 3 answer to the units asked for

1. The reaction is not balanced correctly as written. It is element-balanced but not charge-balanced. It has a total charge of +1 on the left and +2 on the right. The correct equation is

Cu (s) + 2 Ag⁺ (aq) ––�?gt; Cu²⁺ (aq) + 2 Ag (s)

Now it has the same charge on both sides, +2, and is still element-balanced.

Replacing Cu with Mg, the reaction is

Mg (s) + 2 Ag⁺ (aq) ––�?gt; Mg²⁺ (aq) + 2 Ag (s)

2. The "given" substance is the 1 gram of silver. Convert to moles of silver. Moles of silver = grams of Ag / atomic weight of Ag.

3. The "desired" substance is magnesium. How many moles of Mg must react in order to form your number of moles of silver calculated in step 2? Use the coefficients in the balanced reaction. Notice how the units cancel correctly to the desired unit, moles of Mg, in the numerator:

moles of Mg = Your moles of Ag X 1 mol Mg =

1 2 mol Ag

4. Convert moles of Mg to grams. Grams of Mg = moles of Mg X atomic weight of Mg.

And that's it! This is a general approach to working these kinds of problems.

Steve

I need some help with a lab write up please:

T/oC	sample weight/g	NaOH vol./cm3	benzoic acid content/mol	benzoic acid/g	water content/g	water content/mol
58	7.7062	7.6
58	4.0485	5.55
50	19.0906	12.75
50	18.0660	13.55
42	6.36	3.84
42	4.89	2.75
23	14.438	5.3
23	13.2212	4.55

Do you know the molarity of the NaOH solution?

1) Volume of NaOH in liters times molarity equals moles of NaOH = moles of benzoic acid since the reaction has a one-to-one stoichiometry.

2) grams of benzoic acid in the sample = moles X 122.12134 g/mol.

3) grams of water in the sample = total sample weight �?nbsp;grams of benzoic acid

4) moles of water = grams of water / 18.0153 g/mol.

Lastly, what is the significance of the temperatures? As a test, I took the first two samples, each at 58°, and converted to moles of benzoic acid by dividing the sample weights by 122.12134 g/mol (which assumes they are pure benzoic acid). Since these would be the moles of NaOH also, I divided each by the volume of NaOH in liters used in the titration to get NaOH molarities of 8.3030 M and 5.9732 M, respectively. If the percent water was the same in both samples, the calculated molarities should also have been the same, even though incorrect (these are very high molarities, so I expect there is actually a considerable amount of water in these samples). So it looks to me like the same temperature does not equate to the same percent water content.

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