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Study Problems

Experiment 3:

Formula Weight of an Ionic Compound

> 1.- 5.1 g of a carbonate of an unknown metal M, with general formula MxCO3, was thermally decomposed according to the following reaction:

> MxCO3 (s) MxO (s) + CO2 (g)

> The gaseous CO2 was bubbled through a saturated aqueous solution of barium hydroxide (Ba(OH)2 ), and 9.4 g of barium carbonate (BaCO3 )were obtained according to

> CO2 (g) + Ba(OH)2 (aq)BaCO3 (s) + H2O

> a) What's the molecular weight of MxCO3 ?

> Solution:

> First, check to see that the above equations are balanced. They are.

> Second, calculate the formula weight of BaCO3.

> FW(BaCO3) = 137.327 g/mol Ba + 12.011 g/mol C + 3(15.9994 g/mol O)

> FW(BaCO3) = 197.3362 g/mol BaCO3

> Now we can calculate how many moles of BaCO3 were produced.

> = .04763 mol BaCO3

> Since CO2 is in a 1:1 relationship with BaCO3 for which we just calculated the number of moles produced, we state that .04763 mol of CO2 were consumed. If we assume that none of the CO2 produced originally was lost, then we can state that in the first reaction, we produced .04763 moles of CO2 . Note that this is a very unrealistic assumption (a 100% yield does not exist-ideal). Following the same logic, we must have started the whole process off with .04763 mol MxCO3 . since it is in a 1:1 relationship with CO2 as well.

> At this point, we can start our last step in obtaining the formula weight of MxCO3, setting up a proportion. In it, we compare the number of grams in a fraction of a mole of MxCO3 to the number of grams in a whole mole, thus obtaining the formula weight.

> ? = 107.08 g/mol MxCO3

> There is a shortcut we can take with little difficulty, because all of our yields are 100%. Set the two equations up like algebraic equations, add them together, and cross out like terms.

> MxCO3 (s)MxO (s) + CO2 (g)

> + CO2 (g) + Ba(OH)2 (aq)BaCO3 (s) + H2O

> MxCO3 (s) + CO2 (g) + Ba(OH)2 (aq) MxO (s) + CO2 (g)+ BaCO3 (s) + H2O

> Rewrite.

> MxCO3 + Ba(OH)2 (aq) MxO (s) + BaCO3 (s) + H2O

> Now it's a lot easier to see that if our product has .04763mol BaCO3 and our yield is 100%, then we must have begun with .04763mol MxCO3 . We can carry on from there as above to get the formula weight.

> b) What is x and what metal is M?

> Solution:

> How much does a mole of CO3 weigh?

> FW (CO3) = wt (C) + [3 x wt(O)]

> FW (CO3) = 12.011 g/mol C + 3(15.9994 g/mol O) = 60.0092g

> Now we subtract this from the formula weight of MxCO3 in order to obtain the weight of Mx.

> 107.08 g - 60.0092 g = 47.1 g/mol Mx

> Now we have the challenging task of deciding what x is and thus identity of M.

> If x =1, that is 47.1g is the weight of one mole of our metal, not the weight of two or three moles (x=2 or x=3, respectively), then the atomic weight of our metal is 47.1 g/mol. The metal closest to having this weight is Titanium, Ti. (Why did I choose x to be 1 for my first trial? It's really just a guess, but experience tells me that the formula is more likely to be MCO3 than M200 CO3. Also, 1 is the easiest number with which to work. )

> Now we have to ask ourselves if the observed oxidation states of Ti will fit with x =1. If they don't, x doesn't equal 1 and we are not dealing with titanium. According to the CRC Handbook of Chemistry and Physics, the possible oxidation states of Ti are 2+, 3+, and 4+ . Suppose Ti is in its 2+ oxidation state (again, this is just a guess). We know the CO3 anion has a charge of 2- and that the oxidation state (or number) is equal to the charge. Now we consider the "criss-cross" method used in many high school chemistry courses to predict chemical formulas. We'll label Ti and CO3 with their oxidation numbers as superscripts as is commonly done.

> Ti2+ CO32-

> According to this method, the 2+ superscript on Ti becomes the subscript for CO3 and the 2- superscript on CO3 becomes the subscript for Ti . In other words we bring the superscripts down and switch them. That way our compound is Ti2(CO3)2. The like subscripts cancel out, and we are left with TiCO3 as our empirical formula. TiCO3 seems like a likely candidate: it has x = 1 (which is the case if our metal is titanium) and it has a subscript of 1 on CO3 as in our unknown, Mx(CO3)1 . (The subscript on the CO3 anion is just written explictly here for illustrative purposes.) Therefore, we cannot eliminate Ti on the basis of its oxidation states, because at least one of these states fulfills our requirements.

> However, when we further consult the CRC Handbook, we realize that this compound has never been synthesized or isolated; what this means for us is that this is not the identity of our mystery metal.

> If x = 2, then the atomic weight of our metal is 23.55 g/mol. The metal closest to this is Na, sodium.

> Let's take a quick look at the criss-cross method to see if sodium's oxidation state allows for x = 2 and a subscript of 1 on CO3. The oxidation number of sodium is 1+ and of CO3 is 2- as above.

> Na 1+ CO32-

> When we lower and switch the subscripts as done above, we obtain sodium carbonate, Na2CO3. Sodium carbonate fits our requirments that x = 2 in MxCO3 and that the subscript on CO3 is 1.

> And , according to the CRC, this exists.

> Na2CO3 is a good possibility for our unknown compound, but let's go a step further and see what happens if we set x equal to 3.

> If x =3, our metal has an atomic weight of 15.7 g/mol. This is closest to oxygen which is not a metal at all, so x cannot be equal to 3.

> Therefore, we can safely conclude that x = 2, our metal is Na, and our compound is Na2CO3.

> c) What is the percentage composition of MxCO3 ?

> Solution:

> Let's work with the actual compound, Na2CO3 and recalculate its formula weight.

> FW ( Na2CO3) = 105.99 g/mol (No need to show the calculation of formula weight; a similiar calculation was done in question 1a.)

> % Na = x 100% = 43%

> % C = x 100% = 11%

> % O = x 100% = 45%

> d) What volume would the released CO2 occupy at 25ûC and 760 mmHg? (Assume ideal gas behavior)

> Solution:

> We'll use the ideal gas law here and recall that we're dealing with .04763mol of CO2.

> PV = nRT

> (1 atm)V = (.04763 mol)(.08206 L atm mol-1 K-1)(298.15 K)

> V = 1.2L

> 2.- A hydrate, Na2SO4nH2O, contains 55.19% H2O. If you heat 5.0 g of this compound in the oven until it is completely dry:

> a) What is the amount of anhydrous Na2SO4 that you will obtain?

> Solution:

> We subtract the 55.19% H2O from 100% to find out what percent by weight of the hydrated compound is Na2SO4, sodium sulfate.

> 100% Na2SO4 nH2O - 55.19% H2O = 44.81% Na2SO4

> Multiply our 5g hydrated sample by the percent by weight that Na2SO4 represents according to our previous equation.

> 5g x .4481 = 2.2g Na2SO4

> b) What is n in Na2SO4nH2O? [ MW(Na2SO4 ) = 142.04 g/mol, MW(H2O) = 18.02 g/mol ]

> Solution:

> With how many moles of sodium sulfate are we dealing if we have 2.2g as calculated above?

> MW ( Na2SO4) = 142.04 g/mol

> = .3155 mol Na2SO4

> If 2.2 g of our 5g compound is Na2SO4 , the rest must be water.

> 5g -2.2g = 3.8g H2O

> How many moles of water does that represent?

> MW ( H2O) = 18.022 g/mol

> = 3.063 mol

> Now we can compare the moles of H2O and Na2SO4 by dividing both by the smaller of the two, the .3155 moles. (This is standard operating procedure for finding an empirical formula.)

> 1 mol Na2SO4 : 9.7 mol H2O (about 10)

> Keeping the above ratio in mind, we would record the hydrated formula as Na2SO4¥ 10H2O.

> 3.- After heating 5.000 g of a PURE ionic compound X, a final weight of 4.332 g is obtained. The compound X is either CaCl2¥H2O or CaCl2¥6H2O. Which compound is X? [MW(CaCl2) = 110.98 g/mol, MW(H2O) = 18.02 g/mol].

> Solution:

> This question can be solved in the same manner as 2b.

> The 5.000g sample represents the hydrated weight of our compound. The final weight after heating is the dehydrated weight (just CaCl2 is left, if we have taken the time to dry to a constant weight--ie, we've driven out all the water). Therefore, the weight lost is the weight of water in the compound. We'll subtract the final weight from the initial weight to get the weight of water and then figure out how many moles that represents.

> 5.000g - 4.332g = .668g H2O

> MW ( H2O) = 18.022 g/mol

> = .0371 mol H2O

> How many moles of dehydrated CaCl2 do we have?

> MW(CaCl2) = 110.98 g/mol

> = .03903 mol CaCl2

> Let's divide the number of moles of CaCl2 and H2O by the smaller of the two, .0371 moles, in order to get the simplest ratio between them.

> 1 : 1

> The ratio of moles is 1:1. Therefore the formula is 1CaCl21H2O or more simply CaCl2H2O.

> 4.- 0.400 g of an unknown sulfate Mx(SO4)y is dissolved in water and an excess of BaCl2 is added, producing 0.390 g BaSO4(MW: 233 g/mol).

> a) How many moles of sulfate are there in the sample?

> Solution:

> We'll assume that all the sulfate in the unknown, one of our reactants, is found in BaSO4 , one of our products, as we are not told anything about the yield of the reaction. Therefore, the number of moles of sulfate found in BaSO4 equals the number found in the unknown. First we'll calculate how many moles of BaSO4 we have.

> = .001673 mol BaSO4

> It is fairly obvious that if we have .001673 mol BaSO4 , then we have .001673 moles of sulfate since BaSO4 : SO4 are in a 1:1 relationship. Therefore, the moles of sulfate in the unknown are .001673.

> b) What is the percentage-by-mass of sulfate (%SO4)?

> Solution:

> We know how many moles of SO4 are in our unknown metal. Let's convert that information to a weight measurement.

> MW(SO4) = 32.066g/mol S + 4(15.9994g/mol O) = 96.0636g/mol SO4

> (96.0636g/mol)(.001673 mol) = .16071g SO4

> We were told that we have .400g of unknown, so let's get the mass percent that sulfate represents.

> x 100% = 40.2% SO4

> 5.- 2.500 g of an unknown sulfate compound X is heated in the oven at 200 oC for 2 hours until all the hydration water is eliminated, and 1.600 g. of anhydrous sample are obtained.

> a) Calculate the percent weight content of water in compound X.

> Solution:

> How much water was lost when we heated our unknown in the oven? Obviously, it's the hydrated weight minus the weight after the water had been driven out.

> 2.500g -1.600g = .900g

> What percent does the water represent of the whole sample?

> x 100% = 36.0%

> b) The anhydrous sample from above (1.600 g) is dissolved in 100 mL distilled water and treated with 1 mL of concentrated HCl. 10 mL of 1M BaCl2 are then added and all the SO42- is precipitated as BaSO4. After careful filtration and drying, 2.330 g of BaSO4 are obtained. What is the sulfate percent weight content in the unknown compound X.

> [MW(BaSO4) = 233 g/mol; MW(SO4) = 96 g/mol; MW(H2O) = 18 g/mol]

> Solution:

> First we need to find out how many moles of sulfate are in the 2.330g BaSO4.

> = .01 mol BaSO4

> Therefore, we also have .01 mol sulfate because SO4 and BaSO4 are in a 1:1 relationship. (This is the same logic that was followed in question 4a.)

> How many grams of sulfate does this molar amount represent?

> MW (SO4) = 96 g/mol

> (.01 mol SO4 )(96g/mol SO4) = .96g SO4

> We are told that no sulfate is lost in the conversion of our unknown and barium chloride to barium sulfate. Therefore, there is .96g sulfate in the unknown.

> Therefore, let's see what percent by weight that sulfate represents in X.

> x 100% = 38.4% sulfate by weight in X

> (Note: It is not really necessary to specifically state that the percent is by weight; that is generally assumed unless otherwise indicated.)

> c) Use the information from parts (a) and (b) to calculate the percent weight metal content in the unknown compound and to determine the atomic weight of the metal. (Assume that M is a 2+ cation and that the unknown compound has a general formula MSO4nH2O).

> Solution:

> The percent water in X is 36.0 and the percent sulfate is 38. Add these two percentages together.

> 36% + 38.4% = 74.4%

> If 74% of X is water and sulfate, then the remaining 25.6% must be metal.

> Then the question is, how many grams does 25.6% of 2.500g represent?

> .256 x 2.500g = .64g unknown metal

> Suppose we take the formula of our sulfate to be Mx (SO4 )y. It is a lot easier here than in 1b to determine the value of x, because the question also tells us that M is a 2+ cation. Therefore, M is in a 2+ oxidation state. We know sulfate is an anion with a charge of 2- and thus has an oxidation number of 2-.

> M2+ SO42-

> If we use the criss-cross method as above, we obtain the formula the formula of MSO4 that the question indicates. From 5b, we know we have .01mol of SO4, therefore, examining the formula that we just derived, we can see there must be .01 mol M 2+ as well. The logic is the same as question 4b: M 2+ and SO4 are in a 1:1 relationship, so if SO4 is present in .01 mol, then M 2+ is as well.

> Then, if .64g M 2+ is .01 mol, how many grams are in one mole of M2+?

> ? = 64 g per mole of M 2+

> What element has an atomic weight (AW) of 64g?

> Cu

> Notice the answer is very close to being zinc also. Can either answer be disqualified? Both exist as 2+ ions. Both zinc sulfate and copper sulfate exist. Perhaps we can eliminate one of them on the basis of the hydrates that they form. Let's examine how many moles of water are associated with 1 mole of our unknown, ie we want to discover the complete formula, including the value of n.

> How many moles of water are in the .900g of H20 in our hydrate?

> MW(H2O) = 18 g/mol

> = .05mol H2O

> Let's restate how many moles we have of H2O, SO42-, and M2+.

> .05mol H2O .01 mol SO4 2- .01 mol M 2+

> As we do any time we're trying to find an empirical formula, we divide the moles of each by the component with the least number of moles.

> 5mol H2O : 1mol SO4 : 1mol M

> Therefore, MSO4 5H2O

> If we consult the CRC Handbook of Chemistry and Physics, we see that only copper sulfate exists in the pentahydrate form. Zinc sulfate appears as a hexahydrate and heptahydrate. Therefore, we can finally identify the compound as CuSO4 5H2O. You could have easily just accepted the metal to be copper as our atomic weight was slightly closer to it than to zinc, but it's a good idea to familiarize yourself with the CRC. It is probably the most important chemistry reference book.