Week 3 July 22, 2016 Worksheet. Review III. 1 mol = a) 10.0 g Na b) 10.0 g Fe c) 10.0 g Ag d) 10.0 g Cu e) 10.0 g Ba

Transcription

1 Review III = Which sample contains the greatest number of atoms? a) 10.0 g Na b) 10.0 g Fe c) 10.0 g Ag d) 10.0 g Cu e) 10.0 g Ba 10.0 g Na 10.0 g Fe Na g Na atoms = atoms Na Fe g Fe atoms = atoms Fe Ag 10.0 g Ag g Ag atoms = atoms Ag 10.0 g Cu Cu g Cu atoms = atoms Cu Ba 10.0 g Ba g Ba atoms = atoms Ba Another way to think about this problem (without actually doing any calculations) is to think about the conversion. In all cases we start with 10.0 g of an element, and then we divide by the mass of each element (blue numbers), and then multiply by Avogadro s number. So the only difference between each calculation is the mass (blue numbers) that we use to divide 10.0 g, so since Na has the lightest mass, the answer (conversion) will be the largest since we are dividing by the smallest number. 2. How many grams of carbon are in a g sample of carbon suboxide (C 3 )? a) g b) g c) g d) g e) g First, find the molar mass of C 3 : C 3 = 3 mol C g g C + 2 mol O O = g C O 3 2 Now we can imagine the following reaction: 3C + 2O 1C 3 which tells us that for every of C 3, we have 3 mol of C and 2 mol O. So we can first convert the g C 3 into moles of C 3, and then into moles of C using the mole ratio, and then into grams of C g C 3 1mol C O mol C g C g C 3 1mol C 3 C = g C

2 3. A compound was analyzed to contain 79.89% carbon and 20.11% hydrogen. Determine the empirical formula of the compound. a) CH b) CH 2 c) CH 3 d) CH 4 e) C 7 H 20 Step 1: Convert the percentages into masses (assume you have a 100 g sample): g C g H Step 2: Convert these masses into moles: g C 1mol C 1mol H = mol C g H = H g C 1.01 g H Step 3: Divide both mole amounts by the smaller number (6.652): mol C = C mol H = 2.99 mol H Step 4: Write the empirical formula based on these numbers (2.99 3): C 1 H 3 4. How many grams of aluminum sulfate are in a mol sample? a) 77.5 g b) 152 g c) 172 g d) 185 g e) 216 g First, write the chemical formula for aluminum sulfate: Al 2 (SO 4 ) 3. Then find the molar mass of Al 2 (SO 4 ) 3 : 2 mol Al g g g Al + 3 mol S S + 12 mol O O = g Al (SO ) Now we can convert mol Al 2 (SO 4 ) 3 to grams: mol Al 2 (SO 4 ) g Al 2 (SO 4 ) 3 Al 2 (SO 4 ) 3 = 216 g Al 2 (SO 4 ) 3 5. If you have equal mole samples of each of the following compounds, which compound contains the greatest number of oxygen atoms? a) Magnesium Nitrate Mg(NO 3 ) 2 6 O atoms b) Dinitrogen Pentoxide N 2 O 5 5 O atoms c) Iron (III) Phosphate FePO 4 4 O atoms d) Barium Oxide BaO 1 O atoms e) Potassium Acetate KC 2 H 3 2 O atoms If we write out the chemical formula for each compound, we can easily count the number of oxygen atoms for each compound. For instance, H 2 O has 1 O atom. The chemical formulas and O atoms are given in blue next to each compound name.

3 6. How many of the following pairs of compounds have the same empirical formula? I. C 2 H 2 and C 6 H 6 CH and CH II. C 2 H 6 and C 4 H 10 CH 3 and C 2 H 5 III. N and N 2 O 4 N and N IV. C 12 H 10 O and C 6 H 6 O C 12 H 10 O and C 6 H 6 O a) 0 b) 1 c) 2 d) 3 e) 4 7. If you have water molecules present in a sample, what must be true? I. e of water is present in the sample True, because we know that = II g of water is present in the sample True. If we have of H 2 O, then: 1mol H 2 O = 2 mol H g g 1mol H + O 1mol O = g III atoms are present in the sample True. We know that for every ecule of H 2 O, we have 2 atoms of H and 1 atom of O. So we every ecule of H 2 O has 3 atoms. 3 ( molecules H 2 O) = atoms IV g of hydrogen are present in the sample True, because H 2 O has 2 mol H and O, and 2 mol H weighs g. V. 2 moles of hydrogen are present in the sample True, see IV. a) I only b) I, II, IV c) I, II, V d) II, IV e) All 8. How many moles are in a 50.0 g sample of ammonium carbonate? a) mol b) mol c) 0.53 d) 0.64 e) 3.13 mol First, write the chemical formula for ammonium carbonate: (NH 4 ) 2 CO 3. Then find its molar mass: 2 mol N g 1.01 g g g N + 8 mol H H + C C + 3 mol O O = g Now we can convert 50.0 g (NH 4 ) 2 CO 3 to moles: 50.0 g (NH 4 ) 2 CO 3 (NH 4 ) 2 CO g (NH 4 ) 2 CO 3 = mol (NH 4 ) 2 CO 3

4 9. The percent by mass of nitrogen is 46.7% for a species containing only nitrogen and oxygen. Which of the following could be this species? a) N 2 O 5 b) N 2 O c) N d) NO e) NO 3 Because 46.7% of the compound is nitrogen, we know that the remainder must be oxygen: 100% 46.7% = 53.3% O. Now we can follow the steps from problem 3. Step 1: Convert the percentages into masses (assume you have a 100 g sample): 46.7 g N 53.3 g O Step 2: Convert these masses into moles: 46.7 g N 1mol N 1mol O = mol N 53.3 g O = 3.33 O g N g O Step 3: Divide both mole amounts by the smaller number (3.331): mol N = N 3.331mol O = O Step 4: Write the empirical formula based on these numbers: NO 10. A compound containing only sulfur and nitrogen is 69.6% S by mass. The molar mass is g/mol. What is the correct name for this compound? a) Tetrasulfur Dinitride b) Trisulfur Hexanitride c) Sulfur Mononitride d) Tetrasulfur Tetranitride e) Disulfur Hexanitride Because 69.6% of the compound is sulfur, we know that the remainder must be nitrogen: 100% 69.6% = 30.4% O. Now we can follow the steps from problem 3/9. Step 1: Convert the percentages into masses (assume you have a 100 g sample): 69.6 g S 30.4 g N Step 2: Convert these masses into moles: 69.6 g S 1mol S 1mol N = mol S 30.4 g N = mol N g S g N Step 3: Divide both mole amounts by the smaller number (2.170): mol S mol N = S = N Step 4: Write the empirical formula based on these numbers: SN Step 5: Name the compound, which is covalent: sulfur mononitride

5 11. Using solubility rules, predict which of the following pairs will produce a precipitate. a) NH 4 Cl and Na 2 S (NH 4 ) 2 S (aq) and NaCl (aq) b) HNO 3 and CuSO 4 H 2 SO 4 (aq) and CuNO 3 (aq) c) FeCl 2 and KOH Fe(OH) 2 (s) and KCl (aq) d) NiCl 2 and (NH 4 ) 2 SO 4 NiSO 4 (aq) and NH 4 Cl (aq) e) All of the above False. You can determine the product of each reaction by either using a chart with the cations/anions, or swapping the two ions in each reaction. Then you want to use solubility rules to determine which products are soluble or insoluble (solid or precipitate). The products and their solubility are written next to each pair of reactants. The only insoluble (solid) compound is Fe(OH) The correct net ionic equation for the reaction between nickel (II) chloride and sodium sulfide is: a) Ni (aq) + S (aq) NiS (s) b) NiCl 2 (aq) + Na 2 S (aq) NiS (s) + 2NaCl (aq) c) Ni 2+ (aq) + S 2 (aq) NiS (s) d) Na + (aq) Cl (aq) NaCl (aq) e) Ni 2+ (aq) + Cl (aq) + Na + (aq) + S 2 (aq) NiS (s) + NaCl (aq) First determine the chemical formulas for nickel (II) chloride and sodium sulfide, which are the reactants: NiCl 2 (aq) + Na 2 S (aq) These are both soluble so they will exist as ions in solution: Ni 2+ (aq) + Cl (aq) + Na + (aq) + S 2 (aq) Now you want to determine the products of this reaction. You can either set up a chart with the cations/anions or swap the two ions. Use the solubility rules to determine the solubility of each of the products as aqueous (soluble) or solid (insoluble): Ni 2+ (aq) + Cl (aq) + Na + (aq) + S 2 (aq) NaCl (aq) + NiS (s) In a net ionic equation we only care about the precipitate (solid) and the ions that form it. The others (Na + and Cl ) are spectactor ions and do not react. We can now write: Ni 2+ (aq) + S 2 (aq) NiS (s) 13. Which of the following compounds is soluble? a) Lead (II) Sulfate b) Magnesium Sulfate c) Iron (III) Phosphate d) Nickel (II) Hydroxide e) Magnesium Hydroxide

6 14. According to the following unbalanced chemical reaction 1C 12 H 22 O 11 (s) + 12 (g) 12C (g) + 11H 2 O (g) How many moles of oxygen gas are required to react completely with 2.0 moles of C 12 H 22 O 11 (s)? a) 12 mol b) 24 mol c) 30 mol d) 35 mol e) Other After we have balanced the chemical equation (hint: start by trying to balance the number of oxygens), we see that for every e of C 12 H 22 O 11, 12 moles of are needed for the reaction. We can use mole ratios to convert now: 12 mol O 2.0 mol C 12 H 22 O 11 2 = 24 mol O C 12 H According to the following unbalanced chemical reaction 1CH C + 2H 2 O If 5.00 g of CH 4 is burned, what mass of water can be produced? a) 5.62 g b) 10.0 g c) 11.2 g d) 18.0 g e) 36.0 g After we have balanced the chemical equation (hint: start by trying to balance the number of hydrogens), we see that for every e of CH 4, 2 moles of H 2 O are produced. We can now convert 5.00 g CH 4 into moles of CH 4 using its molar mass (16.05 g), then use mole ratios to convert to moles of water, and then convert moles of H 2 O into grams using its molar mass (18.02 g): 5.00 g CH 4 CH g CH 4 2 mol H 2 O CH g H 2 O H 2 O = 11.2 g H 2 O