Feb. 9, 2005

 Solutions are in red.

1.Complete the following table. The first row is done as an example.

2. Name the chemical or give the chemical formula for the following:

(a) CuI: copper(I) iodide or cuprous iodide

(b) NH₄NO₃: ammonium nitrate

(c) tin(II) fluoride: SnF₂

(d) chloride monoxide: Cl₂O

(a) CuI₂: copper(II) iodide or cupric iodide

(b) Sr₃N₂: strontium nitride

(c) chromium(III) carbonate: Cr(III)₂(CO₃)₃

(d) chlorine trifluoride: ClF₃

(a) CoI₂: cobalt(II) iodide

(b) NaHCO₃: sodium bicarbonate or sodium hydrogencarbonate

(c) copper(II) nitrate: Cu(NO₃)₂

(d) diphorphorous pentaoxide: P₂O₅

3. A concerned and scientificly literate mouse finds some crumbs of a Quaker Chewy Granola bar (Butterfinger pieces). The little furry mammal hops up to the cupboard and examines the nutritional information on the back of the Quaker Chewy Granola bar box before eating the crumbs. The mammal finds that the suggested serving size is 1 bar (28 g) and that one serving contains 110 Calories. Now a food Calorie (note that the "c" in food Calories is capitalized) is actual 1,000 chemical calories (note that the "c" in chemical calories is not capitalized) and 1 chemical calorie is exactly equal to 4.184 Joules. How many joules of energy will this little guy get if he eats a crumb that weighs 17 mg 25 mg 67 mg?

28 g = 110 (food) Calories

1 Calorie = 1,000 (chemical) calories

1 calorie = 4.184 J

(a) so for 17 mg to Joules:

17 mg(1g/1,000 mg)(110 Calories/28 g)(1,000 calories/1 Calorie)(4.184 J/1 calorie) = 279 J rounded to 280 J or 2.8 x 10² J

(b) so for 25 mg to Joules:

25 mg(1g/1,000 mg)(110 Calories/28 g)(1,000 calories/1 Calorie)(4.184 J/1 calorie) = 411 J rounded to 410 J or 4.1 x 10² J

(c) so for 67 mg to Joules:

67 mg(1g/1,000 mg)(110 Calories/28 g)(1,000 calories/1 Calorie)(4.184 J/1 calorie) = 1,101 J rounded to 1,100 J or 1.1 x 10³ J

4. If the nucleus of an atom had a diameter of 1 cm 1 mm 1 m, what would be the approximate diameter of the atom?

(a) 1 cm

(b) 10 cm

(c) 1 m

(d) 10 m

(e) 100 m

(f) 1 km

(g) 10 km

(h) 100 km

(i) 1,000 km

The nucleus of an atom is on the order of fentometers (1 x 10^-15 meters) and the size of the atom, which includes the electron cloud, is on the order of angstroms (1 x 10^-10 meters). So, the diameter of the atom is 5 orders of ten larger than the nucleus.

If Nucleus is 1 cm (1 x 10^-2 m) then the diameter of the atom would be 1 km (1 x 10³ m).

So for 1 cm, the answer is 1 km.

If Nucleus is 1 mm (1 x 10^-3 m) then the diameter of the atom would be 100 m (1 x 10² m).

So for 1 mm, the answer is 100 m.

If Nucleus is 1 m then the diameter of the atom would be 100 km (1 x 10⁵ m).

So for 1 m, the answer is 100 km.

5. The density of an organic compound with the molecular formula C₆H₁₂O is 0.879 g/mL. How many (a number not a number of moles) carbon atoms are there in 81.9 mL of this compound?

molecular weight of C₆H₁₂O is 100. g/mol

81.9 mL(0.879g/mL)(1 mol C₆H₁₂O/100. g)(6.02 x 10²³ molecules/1 mol)(6 atoms C/1 molecule) = 2.60 x 10²⁴ C atoms

5. The density of an organic compound with the molecular formula C₈H₁₆O is 0.789 g/mL. How many (a number not a number of moles) carbon atoms are there in 18.9 mL of this compound?

molecular weight of C₈H₁₆O is 128. g/mol

18.9 mL(0.789g/mL)(1 mol C₈H₁₆O/128. g)(6.02 x 10²³ molecules/1 mol)(8 atoms C/1 molecule) = 5.61 x 10²³ C atoms

5. The density of an organic compound with the molecular formula C₁₀H₁₈O is 0.978 g/mL. How many (a number not a number of moles) carbon atoms are there in 91.8 mL of this compound?

molecular weight of C₁₀H₁₈O is 154. g/mol

91.8 mL(0.978g/mL)(1 mol C₆H₁₂O/154. g)(6.02 x 10²³ molecules/1 mol)(10 atoms C/1 molecule) = 3.51 x 10²⁴ C atoms

6. A chemist analyzed a 0.351 g sample of an unknown organic compound that contains carbon, hydrogen and oxygen. The chemist completely burned the sample in an excess of oxygen gas yielding 0.799 g of carbon dioxide and 0.327 g of water. (All the carbon in the original sample ended up in the carbon dioxide and all the hydrogen ended up in the water.) A low resolution mass spectrum of the compound indicated that the molecular weight of the compound was 116 +/- 2 amu. Calculate the chemical formula of the organic compound.

moles of CO₂ to moles of carbon in the sample: 0.799g (1 mol CO₂/44.0 g CO₂)(1 mol C/1 mol CO₂) = 0.018159 mol of C

moles of H₂O to moles of hydrogen in the sample: 0.327g (1 mol H₂O /18.0 g H₂O )(2 mol H/1 mol H₂O ) = 0.036333 mol of H

Now, how much oxygen was in that original sample? We know the moles of carbon (therefore we know the weight of carbon) and we know the moles of hydrogen (therefore we also know its weight) in the original sample. The rest of the weight must be oxygen. So convert 0.018159 mol of C to grams add that to grams from 0.036333 mol of H and subtract that from the original weight (0.351 g):

0.018159 mol of C(12.0 g C/1 mol C) = 0.218 g C

0.036333 mol of H (1.0 g H/1 mol H) = 0.0363 g H

0.351g - 0.254 g (C and H) = 0.097 g of O convert this to moles: 0.097 g O(1 mol O/16 g O) = 0.00604 mol O

Did gives the empirical formula of : C_0.0182H_0.0363O_0.00604 gives C₃H₆O₁ with a formula weight of 58.

The mass spectrum of the compound is 116 +/- 2 amu so C₃H₆O₁ isn't the molecular formula.

Try 2(C₃H₆O₁ ) or C₆H₁₂O₂ which gives formula weight of 2 x 58 = 116.

ANSWER is C₆H₁₂O₂

6. A chemist analyzed a 0.351 g sample of an unknown organic compound that contains carbon, hydrogen and oxygen. The chemist completely burned the sample in an excess of oxygen gas yielding 0.757 g of carbon dioxide and 0.310 g of water. (All the carbon in the original sample ended up in the carbon dioxide and all the hydrogen ended up in the water.) A low resolution mass spectrum of the compound indicated that the molecular weight of the compound was 103 +/- 2 amu. Calculate the chemical formula of the organic compound.

moles of CO₂ to moles of carbon in the sample: 0.757g (1 mol CO₂/44.0 g CO₂)(1 mol C/1 mol CO₂) = 0.01720 mol of C

moles of H₂O to moles of hydrogen in the sample: 0.310g (1 mol H₂O /18.0 g H₂O )(2 mol H/1 mol H₂O ) = 0.03444 mol of H

Now, how much oxygen was in that original sample? We know the moles of carbon (therefore we know the weight of carbon) and we know the moles of hydrogen (therefore we also know its weight) in the original sample. The rest of the weight must be oxygen. So convert 0.01720 mol of C to grams add that to grams from 0.03444 mol of H and subtract that from the original weight (0.351 g):

0.01720 mol of C(12.0 g C/1 mol C) = 0.206 g C

0.03444 mol of H (1.0 g H/1 mol H) = 0.0344 g H

0.351g - 0.241 g (C and H) = 0.110 g of O convert this to moles: 0.110 g O(1 mol O/16 g O) = 0.00689 mol O

Did gives the empirical formula of : C_0.0172H_0.0344O_0.00689 gives C_2.5H₅O₁ with a formula weight of 51.

The mass spectrum of the compound is 103 +/- 2 amu so C_2.5H₅O₁ isn't the molecular formula.

Try 2(C_2.5H₅O₁ ) or C₅H₁₀O₂ which gives formula weight of 2 x 51 = 102.

ANSWER is C₅H₁₀O₂

6. A chemist analyzed a 0.351 g sample of an unknown organic compound that contains carbon, hydrogen and oxygen. The chemist completely burned the sample in an excess of oxygen gas yielding 0.832 g of carbon dioxide and 0.340 g of water. (All the carbon in the original sample ended up in the carbon dioxide and all the hydrogen ended up in the water.) A low resolution mass spectrum of the compound indicated that the molecular weight of the compound was 129 +/- 2 amu. Calculate the chemical formula of the organic compound.

moles of CO₂ to moles of carbon in the sample: 0.832g (1 mol CO₂/44.0 g CO₂)(1 mol C/1 mol CO₂) = 0.01891 mol of C

moles of H₂O to moles of hydrogen in the sample: 0.340g (1 mol H₂O /18.0 g H₂O )(2 mol H/1 mol H₂O ) = 0.03777 mol of H

Now, how much oxygen was in that original sample? We know the moles of carbon (therefore we know the weight of carbon) and we know the moles of hydrogen (therefore we also know its weight) in the original sample. The rest of the weight must be oxygen. So convert 0.01891 mol of C to grams add that to grams from 0.03777 mol of H and subtract that from the original weight (0.351 g):

0.01891 mol of C(12.0 g C/1 mol C) = 0.227 g C

0.03777 mol of H (1.0 g H/1 mol H) = 0.0378 g H

0.351g - 0.265 g (C and H) = 0.086 g of O convert this to moles: 0.086 g O(1 mol O/16 g O) = 0.00539 mol O

Did gives the empirical formula of : C_0.0189H_0.0377O_0.00539 gives C_3.5H₇O₁ with a formula weight of 65.

The mass spectrum of the compound is 129 +/- 2 amu so C_3.5H₇O₁ isn't the molecular formula.

Try 2(C_3.5H₇O₁ ) or C₇H₁₄O₂ which gives formula weight of 2 x 65 = 130.

ANSWER is C₇H₁₄O₂

7. A brave chemist threw 12.3 g of lithium metal into a vessel containing 109 g of chlorine gas. An immediate violent reaction occurred, generating the salt, lithium chloride. (Don't try this at home!) What is the maximum amount of lithium chloride, in grams, that can be made in this reaction? For full credit you must:

(a) write a balanced chemical equation and

(b) determine the limiting and excess reagents, if they exist.

(a) 2Li(s) + Cl₂(g) ----> 2LiCl(s)

(b) Calculate the amount of the reactants and then determine limiting reagent:

12.3 g Li(1 mol Li/6.94 g Li) = 1.77 mol Li.

How much chlorine do we need to consume all this lithium?

1.77 mol Li (1 mol Cl₂/2 mol Li) = 0.886 moles of chlorine needed to react with all the lithium. Do we have enough chlorine to do this?

109 g Cl₂(1 mol Cl₂/70.9 g Cl₂) = 1.54 mol Cl₂ .

How much lithium do we need to consume all this chlorine?

1.54 mol Cl₂ (2 mol Li/1 mol Cl₂) = 3.07 moles of lithium needed to react with all the chlorine. We don't have enough lithium to consume all this chlorine!

So which is the limiting reagent? Lithium is limiting, so base our product calculation on the amount of lithium - the limiting reagent.

1.77 mol Li (2 mol LiCl/2 mol Li) = 1.77 mol LiCl

convert to grams to get the maximum yield of lithium chlorine:

1.77 mol LiCl (42.4 g LiCl/1 mol LiCl) = 75.0 g of LiCl possible

7. A brave chemist threw 12.3 g of sodium metal into a vessel containing 109 g of chlorine gas. An immediate violent reaction occurred, generating the salt, sodium chloride. (Don't try this at home!) What is the maximum amount of sodium chloride, in grams, that can be made in this reaction? For full credit you must:

(a) write a balanced chemical equation and

(b) determine the limiting and excess reagents, if they exist.

 (a) 2Na(s) + Cl₂(g) ----> 2NaCl(s)

(b) Calculate the amount of the reactants and then determine limiting reagent:

12.3 g Na(1 mol Li/23.0 g Na) = 0.535 mol Na.

How much chlorine do we need to consume all this lithium?

0.535 mol Na (1 mol Cl₂/2 mol Na) = 0.267 moles of chlorine needed to react with all the sodium. Do we have enough chlorine to do this?

109 g Cl₂(1 mol Cl₂/70.9 g Cl₂) = 1.54 mol Cl₂ .

How much sodium do we need to consume all this chlorine?

1.54 mol Cl₂ (2 mol Na/1 mol Cl₂) = 3.07 moles of sodium needed to react with all the chlorine. We don't have enough lithium to consume all this chlorine!

So which is the limiting reagent? Sodium is limiting, so base our product calculation on the amount of sodium - the limiting reagent.

0.535 mol Na (2 mol NaCl/2 mol Na) = 0.535 mol NaCl

convert to grams to get the maximum yield of sodium chlorine:

0.535 mol NaCl (58.4 g NaCl/1 mol NaCl) = 31.2 g of NaCl possible

7. A brave chemist threw 12.3 g of potassium metal into a vessel containing 109 g of chlorine gas. An immediate violent reaction occurred, generating the salt, potassium chloride. (Don't try this at home!) What is the maximum amount of potassium chloride, in grams, that can be made in this reaction? For full credit you must:

(a) write a balanced chemical equation and

(b) determine the limiting and excess reagents, if they exist.

  (a) 2K(s) + Cl₂(g) ----> 2KCl(s)

(b) Calculate the amount of the reactants and then determine limiting reagent:

12.3 g K(1 mol Li/39.1 g K) = 0.315 mol K.

How much chlorine do we need to consume all this potassium?

0.315 mol K (1 mol Cl₂/2 mol K) = 0.157 moles of chlorine needed to react with all the potassium. Do we have enough chlorine to do this?

109 g Cl₂(1 mol Cl₂/70.9 g Cl₂) = 1.54 mol Cl₂ .

How much potassium do we need to consume all this chlorine?

1.54 mol Cl₂ (2 mol K/1 mol Cl₂) = 3.07 moles of potassium needed to react with all the chlorine. We don't have enough potassium to consume all this chlorine!

So which is the limiting reagent? Potassium is limiting, so base our product calculation on the amount of potassium - the limiting reagent.

0.315 mol K (2 mol LiCl/2 mol K) = 0.315 mol KCl

convert to grams to get the maximum yield of potassium chlorine:

0.315 mol NaCl (74.6 g KCl/1 mol LKCl) = 23.5 g of KCl possible

8.(a) Element X has two stable isotopes, X-128 and X-131. If the atomic weight of X is 128.80, what are the percent abundances of each isotope?

 x(128) + y(131) = 128.80 and x + y = 1 so y = 1 - x and our new equation is:

128x + 131(1 - x) = 128.80, solve for x:

128x - 131x +131 = 128.80

-3x = -2.2 ; x = 0.733 so

X-128 is 73.3% abundant &

X-131 is 26.7% abundant

 (b) If 5.37 g of X combines with 1.00 g of oxygen to make a stable oxide, what is the charge on the X ion?

Convert grams of X and oxygen into moles:

1.00 g O(1 mol O/16.0 g O) = 0.0625 mol O

5.37 g X(1 mol X/128.80 g X) = 0.0417 mol X

gives a formula of X_0.0417O_0.0625 or  X_0.667O₁ or X₂O₃

Since oxide has a -2 charge, X must have a +3 charge

8.(a) Element Y has two stable isotopes, Y-97 and Y-93. If the atomic weight of Y is 94.73, what are the percent abundances of each isotope?

  x(97) + y(93) = 94.73 and x + y = 1 so y = 1 - x and our new equation is:

97x + 93(1 - x) = 94.73, solve for x:

97x - 93x + 93 = 94.73

4x = 1.73 ; x = 0.433 so

Y-97 is 43.3% abundant &

Y-93 is 56.7% abundant

(b) If 2.37 g of Y combines with 1.00 g of oxygen to make a stable oxide, what is the charge on the Y ion?

Convert grams of Y and oxygen into moles:

1.00 g O(1 mol O/16.0 g O) = 0.0625 mol O

2.37 g Y(1 mol Y/94.73 g Y) = 0.0250 mol Y

gives a formula of Y_0.0250O_0.0625 or  Y₁O_2.5 or Y₂O₅

Since oxide has a -2 charge, Y must have a +5 charge  

8.(a) Element Z has two stable isotopes, Z-58 and Z-63. If the atomic weight of Z is 59.97, what are the percent abundances of each isotope?

  x(58) + y(63) = 59.97 and x + y = 1 so y = 1 - x and our new equation is:

58x + 63(1 - x) = 59.97, solve for x:

58x - 63x + 63 = 59.97

-5x = -3.03 ; x = 0.606 so

Z-58 is 60.6% abundant &

Z-63 is 39.4% abundant

(b) If 1.07 g of Z combines with 1.00 g of oxygen to make a stable oxide, what is the charge on the Z ion?

 Convert grams of Z and oxygen into moles:

1.00 g O(1 mol O/16.0 g O) = 0.0625 mol O

1.07 g Z(1 mol Z/59.97 g Z) = 0.0178 mol Z

gives a formula of Z_0.0178O_0.0625 or  Z₁O_3.5 or Z₂O₇

Since oxide has a -2 charge, Z must have a +7 charge