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Synthesis of Calcium Phosphate: Jumping Through the Limiting Reagent
Hoop.
Objective: Synthesize calcium phosphate from a mixture of calcium chloride and sodium phosphate.
Determine which reagent is limiting.
Quantitatively determine the % composition of the original mixture.
Introduction:
In your career as a scientist, you will, when you do any analytical or synthetic work, ask the following
types of questions:
1. “If I start with this much stuff, how much product will I be able to make?”
2. “I’ve got a bucket full of stuff here. What stuff is in the bucket, and how much?”
You have been asking yourself this kind of question ever since you were old enough to see over the kitchen
counter. “I have 3 slices of bologna, (or, if you were brought up by my mother, textured tofu cheese-flavored
soy product) and 5 slices of bread. How many sandwiches can I make?” You’d know that every sandwich
requires 1 slice of bologna and 2 slices of bread, Clearly you have enough bread for 2 sandwiches, and enough
bologna for 3 sandwiches. So, how many bologna sandwiches can you make? Two. And you will have
bologna left over. Simple. You knew that. How did you know? You know that when you are making
sandwiches, the starting materials, the bread and bologna have to be put together in a fixed ratio: 1 slice of
bologna to 2 slices of bread. In the language of the scientific tribe, filling and bread must go together in a
stoichiometric ratio. The sandwich problem you just solved is what is known in the trade as a LIMITING
REAGENT problem. The amount of product (sandwich) you can make depends on how much reactant (filling
and bread) you start with. Because the filling and bread come together in a fixed ratio (1:2) the amount of bread
in this case determines how many sandwiches you can make. Bread is the limiting reagent, and bologna is the
excess reagent.
What does this have to do with chemistry?
Well, suppose you perform the following reaction:
1 HCl(aq) + 1 NaOH(aq)  1 H2O(l) + 1 NaCl(aq)
This says that 1 mole of hydrochloric acid reacts with one mole of sodium hydroxide (a 1:1 stoichiometric
ratio), producing 1 mole of water and 1 mole of sodium chloride.
Remember: moles are proportional to NUMBERS OF PARTICLES.
If you started with 1 mole of HCl and 1 mole of NaOH, (reactants in stoichiometric ratio) you’d make 1 mole of
NaCl. There will be no starting material (HCl or NaOH) left over.
If you started with 2 moles of HCl and 1 mole of NaOH, you’d make 1 mole of NaCl. NaOH is the limiting
reagent. There would be 1 mole of HCl left unreacted, and no NaOH left over.
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If you started with 25 moles of HCl and 1 mole of NaOH, you’d still only make 1 mole of NaCl. Again, NaOH
is the limiting reagent. NaOH would be used up. 24 moles of HCl would be left.
25 moles HCl (1 mole NaOH ) = 25 moles NaOH needed to react completely with the HCl. 1 mole NaOH
( 1 mole HCl )
of NaOH is not enough. NaOH is the limiting reagent
Suppose you have a reaction in which the stoichiometric ratio is not 1:1. Consider this reaction:
3CaCl2 (aq) + 2Na3PO4(aq)  Ca3 (PO4)2(s)↓ + 6NaCl(aq)
This says that 3 moles of calcium chloride react with 2 moles of sodium phosphate, and produce 1 mole of
calcium phosphate (the down arrow ↓ says that calcium phosphate is insoluble in water and precipitates).
BEFORE YOU COME TO LAB: WRITE THE IONIC AND NET IONIC EQUATION
FOR THIS REACTION. HAVE THIS ON A SEPARATE PAGE IN YOUR
NOTEBOOK. BE PREPARED TO TURN IN A CARBON COPY.
Suppose you mixed 3 moles of calcium chloride with 2 moles of sodium phosphate. You’d make 1 mole of
calcium phosphate, 6 moles of sodium chloride, and no starting materials would be left over. That’s easy. You
can do it in your head.
Suppose you mixed 2 moles of calcium chloride with 3 moles of sodium phosphate. How many moles of
calcium phosphate would you make?
First, determine the limiting reagent:
2 moles CaCl2 (2 mol Na3PO4 ) = 1.33 moles Na3PO4 is needed to react with 2 moles CaCl2.
(3 moles CaCl2)
Do we have enough? We have 3 moles of Na3PO4, so in fact
we do. Na3PO4 is the excess reagent and CaCl2 is the limiting reagent.
Or you could determine the limiting reagent this way:
3 moles Na3PO4 (3 mol CaCl2 ) = 4.5 moles CaCl2 is needed to react with 3 moles Na3PO4.
(2 moles Na3PO4) Do we have enough CaCl2? We only have 2 moles of CaCl2, so in fact
we DON’T. CaCl2 is limiting and Na3PO4 is in excess.
Now, how many moles of Ca3(PO4)2 can we make with 2 moles of CaCl2? (Always use the limiting reagent).
2 moles CaCl2 (1 mol Ca3(PO4)2 ) = 0.67 moles Ca3 (PO4)2 made.
(3 moles CaCl2)
You’d have no calcium chloride left over, and 3-1.33 = 1.66 moles of Na3PO4 unreacted.
Until now, we have been working with moles of reagents and products. Moles, of course, are directly
proportional to the number of particles reacting, which is what you need to know. In the lab, the way you
measure a reactant or product is to put it on the balance and get a mass. You need to convert from mass (grams)
to moles, and vice versa. You need molecular weights of reactants and products to make the conversion.
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Reagent Table:
Reactant
CaCl2.2H2O
Na3PO4.12H2O
Product
Ca3(PO4)2
NaCl
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MW (g/mole)
147.01
380.12
310.16
58.44
In the problem above, you would be mixing:
294.02 g CaCl2 [2 moles CaCl2 ( 147.01 g ) = 294.02 g CaCl2]
( 1 mol CaCl2)
1140.36 g Na3PO4 [3 moles Na3PO4 ( 380.12 g ) = 294.02 g Na3PO4]
( 1 mol Na3PO4)
and you would obtain:
207.81 g Ca3(PO4)2 [0.67 moles Ca3(PO4)2 ( 310.16 g
) = 207.81 g Ca3(PO4)2]
( 1 mol Ca3(PO4)2)
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BUT WHAT ABOUT REAL LIFE? WHAT ABOUT THIS LAB?
In this particular lab, as often happens in real life, you will be issued an “unknown” mixture of calcium
chloride and sodium phosphate. Your tasks will be:
1. Synthesize calcium phosphate from this mixture;
2. Determine the limiting reagent experimentally;
3. Determine form your data how much calcium chloride and sodium phosphate you started with.
You will need:
EQUIPMENT
LIST
5
2
2
2
1
2
2
2
2
2
2
2
Beakers
Watchglasses
Stirring rods
Bunsen burners
striker
Wire gauze pads
Ring stands
Iron rings
funnels
5” Whatman
qualitative filter
paper
Pastuer pipets &
bulb
Test tubes
Labeling tape
Indelible marker
pencil
250 or 400 mL
to fit on beakers
75 mm
You will also need:
∽ 5 grams of an “unknown” mixture. RECORD THE UNKNOWN NUMBER!
A dropper bottle of 0.5 M CaCl2
On the riser
A dropper bottle of 0.5 M Na3PO4
On the riser
I.
SYNTHESIZE & PURIFY CALCIUM PHOSPHATE FROM A MIXTURE OF
CALCIUM CHLORIDE DIHYDATE AND SODIUM PHOSPHATE
DODECAHYDRATE:
The first thing you need to do is to measure and record in your lab notebook exactly how much unknown you
are starting with.
A. Weigh out your unknown: Label 2 clean (don’t have to be dry) 250 or 400 mL
beakers “trial 1” and “trial 2”. Take the beakers, your unknown, and your lab notebook to the
balance area with you. Write down your unknown number in your notebook! On a piece of
tared weighing paper, weigh out about 2 g (+/- 0.0001 g) of your unknown. Record the mass
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exactly as it reads off the balance in your notebook. Carefully tap the weighed unknown into the
beaker labeled “trial 1”. Weigh out and record another 2 g of unknown (+/- 0.0001 g) and tap it
in the beaker labeled “trial 2”.
Once you know exactly how much stuff you are dealing with, you can start the reaction by adding water.
B. Synthesize Calcium phosphate:
1. Add 200 mL of deionized water to each beaker. Stir the mixture with a stirring rod. You
won’t see a clear, colorless solution. As the calcium chloride and sodium phosphate
dissolve, calcium phosphate is forming and precipitating out. Stir for about a minute, then
leave the stirring rod in the beaker, and cover the beaker with a watch-glass.
☛ Point to ponder…where is the NaCl?
2. Heat the covered beaker over a low flame (do not boil!) for about 10 minutes. See front
bench for set-up. Keep an eye on things…don’t let the solutions boil or spatter.
3. After 10 minutes, turn off the burner flame and let the precipitate settle. You’ve made
calcium phosphate.
Now you need to isolate and purify the calcium phosphate. You need to separate it from water, from sodium
chloride (the other product of the reaction), and from unreacted starting reagents. You will do that by:
 filtering the calcium phosphate (it will hang up in the filter paper, everything else will
wash through),
 washing the calcium phosphate (to get any residual kaka off), and then
 drying the calcium phosphate (to evaporate off the rest of the water).
C. Isolate and Purify Calcium Phosphate:
1. While the precipitate is settling, heat up about 50 mL of deionized water to use as wash water
in step IC9.
2. Get 2 pieces of filter paper, LIGHTLY pencil your initials, your partner’s initials and the trial
number ( “trial 1” or “trial 2”) on the upper edge of the filter papers. Taking them to the
balance area along with your notebook, weigh the filter paper and record the masses (+/0.0001 g)
4. Set up 2 gravity filtration apparati. See front bench for set-up.
5. Label 2 CLEAN (need not be dry) test-tube “trial 1” and “trial 2”.
6. Filter the calcium phosphate: [Remember! the contents of the “trial 1” beaker get filtered
through the funnel with the “trial 1” filter paper]
7. Catch some supernatant from the filtration in your labeled test tubes. Use this to test for the
limiting reagent, in step IIA1.
BEFORE YOU COME TO LAB…What is a supernatant?
8. Finish filtering your calcium phosphate.
9. Rinse the beaker with a small amount (<10 mL) warm deionized wash water to get all
residual calcium phosphate off the glass, and pour it through the filter. Do this twice.
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10. Carefully peel the filter paper out of the funnel and put it in a clean beaker labeled with your
name, your partner’s name, and your lab section. Cover the beakers with watch glasses and
put them in the oven to dry overnight. You now have purified calcium phosphate.
II.
DETERMINE THE LIMITING REAGENT
Now you need to figure out which was the limiting reagent. Since you don’t know how many grams or moles of each reagent you
started with, you must do this experimentally. First, do this thought experiment:
 Supernatant
Ca2+
Cl-
Calcium phosphate ppt.
KNOWN # A
3 mmoles CaCl2 (0.4410 g)
1 mmole Na3PO4 (0.3801 g)
You make up 2 known mixtures. One (A) contains 3 mmoles of CaCl 2 (0.4410 g) and 1 mmole of Na3PO4 (0.3801 g). The limiting
reagent is Na3PO4:
3 mmoles CaCl2(2 mmoles Na3PO4/3 mmoles CaCl2) = 2 mmoles Na3PO4 needed to react completely with the
3 mmoles of CaCl2. We only have 1 mmole Na3PO4.
enough. Na3PO4 is limiting.
Not
If Na3PO4 is limiting, it will all get used up. CaCl2, in excess, will not all get reacted. There will be some dissolved calcium chloride
in the supernatant. If you collect some of the supernatant, and add extra calcium chloride to it, you won’t see a visible change.
However, if you add more sodium phosphate to the supernatant, you will see more cloudy white calcium phosphate form.
3Ca2+(in excess) +
2PO43- (extra
added)
Ca3(PO4)2 ↓
Before you come to lab…In your notebook, determine the limiting and excess reagents in
KNOWN # B which contains 1 mmole CaCl2 (0.1470 g) and 3 mmoles Na3PO4 (1.1480 g). Predict what ions are present in
the supernatant, and what you would expect to see when you add I) additional CaCl 2, and ii) additional Na3PO4.
A1. Using the supernatant you collected in the small test tubes in Step IC7, determine the limiting
reagent. Record in your notebook exactly how you did this, what you saw, and what you concluded.
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III.
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DETERMINE THE % BY MASS OF CalCIUM CHLoride and sodium
phosphate in the unknown mixture.
A1. Retrieve your dried calcium phosphate from the oven, weigh them (+/- 0.0001 g)
and record them in your notebook.
The rest of the lab is calculation.
Set up a spreadsheet as follows:
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Include sample calculations. When using the spreadsheet to do a calculation, write down the formula used.
Make SURE that each cell containing numbers is formatted to give the correct number of significant figures.
Make sure you include the balanced equation!
The calculations are simple..when in doubt, convert to moles!
Suppose you started with 1.2330 g of an unknown. You work it up, test the supernatant, and
notice that you get additional precipitation when you add a couple of drops of 0.5 M CaCl2.
Nothing happens when you add a couple drops of Na3PO4, You conclude that Na3PO4 is in
excess, and CaCl2 is limiting.
You weigh your filter paper and Ca3(PO4)2 and find that you have made 0.0984 g of calcium
phosphate. You want to figure out how many grams of CaCl2 and Na3PO4 you started with.
Well, you know that calcium chloride is limiting. That means ALL the calcium chloride reacted,
and made product.
3CaCl2 (aq) + 2Na3PO4(aq)  Ca3 (PO4)2(s)↓ + 6NaCl(aq)
If you made 1 mole of Ca3 (PO4)2(s) (310.177 g), you’d know you started with 3 moles of
calcium chloride 441.06 g). Well, real life not as easy as whole integers, but not that bad:
When in doubt, convert to moles:
1. Moles Ca3(PO4)2 = 0.0984 g [1 mole Ca3(PO4)2] = 0.000317 mole Ca3(PO4)2
310.177 g
Using the LR:
2. Moles Ca2+ = 0.000317 mole Ca3(PO4)2 [ 3 moles Ca2+ ] = 0.000952 moles Ca2+
1 mole Ca3(PO4)2
(The LR)
3. Mass CaCl2: 0.000952 moles Ca2+ [1 mole CaCl2] [ 147.02 g ]
1 mole Ca2+
mole CaCl2
=
0.140 g
CaCl2
4. %CaCl2.2H20 = 0.140 g CaCl2 x 100 = 11.4% CaCl2
1.2330 g mixture
100 -11.4% = 88.6% Na3PO4.12H2O
If phosphate were limiting, you’d do a similar calculation:
1. Moles Ca3(PO4)2 = 0.0984 g [1 mole Ca3(PO4)2] = 0.000317 mole Ca3(PO4)2
310.177 g
Using the LR:
2. Moles PO43- = 0.000317 mole Ca3(PO4)2 [ 2 moles PO43- ] = 0.000634 moles PO431 mole Ca3(PO4)2
(The LR)
3. Mass Na3PO4: 0.000634 moles PO43- [1 mole Na3PO4] [ 380.12 g ]
1 mole PO43- 1 mol Na3PO4
.
4. %Na3PO4 12H20 = 0.241 g CaCl2 x 100 = 19.6% Na3PO4
=
0.241 g
Na3PO4
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1.2330 g mixture
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