Review Material For Exam I 1. Since 198, a US 1 cent coin is 97.59% zinc and.1% copper. According to the US Mint website, the volume of the coin is 0.070 in 3. It has a density of 6.9 g/cm 3. Calculate the number of copper atoms in the coin. 3 (.5 cm) 6.9 g coin.1 g Cu 1 mole Cu 6.0 x 10 atoms Cu 0.070 in (1 in) 1 cm 100.0 g coin 63.5 g Cu 1 mole Cu = 7.01 x 10 0 atoms Cu. Give the group name of the following: a. I halogen b. Te chalcogen c. Xe noble gas d. Mg alkaline-earth metal e. Li alkali metal 3. Write the formula and proper name of the compound formed by combining: a. Ca + and N 3 Ca 3 N calcium nitride b. K + and O K O potassium oxide c. Fr + and F FrF francium fluoride d. Al 3+ and P 3 AlP aluminum phosphide e. Ba + and Br BaBr barium bromide f. Sr + and Se SrSe strontium selenide. Give the proper name of the following substances (show work for ionic): Na SO SeF Cl O 7 Na O Mn O 3 CuCl Cr O 3 ClF 3 BaS Cu(NO 3 ) N O 5 N F MgI Al (SO ) 3 CaCO 3 sodium sulfate selenium tetrafluoride dichlorine heptoxide sodium oxide manganese(iii) oxide copper(i) chloride chromium(iii) oxide chlorine trifluoride barium sulfide copper(ii) nitrate dinitrogen pentoxide dinitrogen difluoride magnesium iodide aluminum sulfate calcium carbonate
5. Write the chemical formula for the following compounds (show work for ionic): copper(i) oxide Cu O dichlorine pentoxide Cl O 5 tin(ii) fluoride SnF lead(ii) dichromate PbCr O 7 sulfur tetrafluoride SF dinitrogen tetrafluoride N F bismuth(iii) fluoride BiF 3 xenon tetroxide XeO mercury(ii) sulfate HgSO nickel(ii) phosphate Ni 3 (PO ) ammonium nitrate NH NO 3 6. Give the number of protons, neutrons and electrons for the following: p + n o e 58 Cu + 9 9 8 58 Fe 3+ 6 Ne 10 1 10 17 O 8 9 10 187 Au 3+ 79 108 76 7 Br 35 39 36 30 P 3 15 15 18 189 Os + 76 113 7 196 Hg 80 116 80 0 Pb 8 1 8 7. Calculate the following: a. number of atoms in 7.6 g of Li 3 1 mole Li 6.0 x 10 atoms Li 7.6 g Li 6.91 g Li 1 mole Li = 6.7 x 10 3 atoms Li b. number of atoms in 3.0 g of Br 3 1 mole Br 6.0 x 10 molecules Br atoms Br 3.0 g Br 159.808 g Br 1 mole Br 1 molecule Br =.1 x 10 3 atoms Br
c. number of molecules in 3.0 g of NH 3 1 mole NH 6.0 x 10 molecules NH 3.0 g NH 3 17.031 g NH 1 mole NH 3 = 1.5 x 10 molecules NH 3 d. number of molecules in 7.585 g CCl 1 mole CCl 6.0 x 10 molecules CCl 7.585 g CCl 153.83 g CCl 1 mole CCl 3 =.969 x 10 molecules CCl e. number of moles of SO - ions in 1.3 g of Cr (SO ) 3 1 mole Cr (SO ) 3 3 mole SO 1.3 g Cr (SO ) 3 39.175 g Cr (SO ) 1 mole Cr (SO ) = 0.109 mole SO 3 3 f. number of moles of H in 11 g H 3 PO 1 mole H PO 3 mole H 3 11 g H3PO 97.99 g H3PO 1 mole H3PO = 0.3 mole H 8. Oxalic acid is a toxic substance used by laundries to remove rust stains. Its composition is 6.7% C,.0% H and 71.1% O by weight. What is the empirical formula? The formula weight is approximately 90 g/mole. What is the molecular formula? Assume 100 g of compound: 1 mole H.0 g H x.185 mole H/.185 1.000 1.008 g H 1 mole C 6.7 g C x.30 mole C/.185 1.0185 1 1.011 g C 1 mole O 71.1 g O x.0 mole O/.185.036 15.999 g O empirical formula = HCO FW = 5.017 90 g/mole 5.017 g/mole ; molecular formula = x empirical formula = H C O
9. Adipic acid is used in the manufacture of nylon. The composition of the acid is 9.3% C, 6.90% H and 3.8% O by weight. What is the empirical formula? The formula weight is approximately 16 g/mole. What is the molecular formula? Assume 100 g of compound: 1 mole C 9.3 g C x.106 mole C/.7377 1.993 1.5 x 3 1.011 g C = 1 mole H 6.90 g H x 6.85 mole H/.7377.500.5 x 5 1.008 g H = 1 mole O 3.8 g O x.7377 mole O/.7377 1.00 x 15.999 g O = empirical formula = C 3 H 5 O FW = 73.071 16 g/mole 73.071 g/mole ; molecular formula = C 6H 10 O (H C 6 H 8 O ) 10. Butane, C H 10, is used in lighters because it is highly flammable and easily liquified. Calculate the mass of water produced from the complete combustion of 3.00 moles of butane. C H 10 + 13 O 8 CO + 10 H O 10 mole H O 18.015 g H O 3.00 mole C H 10 mole CH10 1 mole HO =.70 x 10 g H O 11. White phosphorous, P, is prepared by fusing calcium phosphate with carbon and sand (SiO ) in an electric furnace. Ca 3 (PO ) + 6 SiO + 10 C P + 6 CaSiO 3 + 10 CO How many grams of calcium phosphate are required to give 5.00 g of phosphorous? 1 mole P mole Ca (PO ) 310.17 g Ca (PO ) 3 3 5.00 g P 13.896 g P 1 mole P 1 mole Ca 3(PO ) = 5.0 g Ca 3 (PO )
1. The following reaction is used to make carbon tetrachloride CS (s) + 3 Cl (g) CCl (s) + S Cl (s) Calculate the number of grams of carbon disulfide needed to react exactly with 6.7 g of chlorine gas. 1 mole Cl 1 mole CS 76.11 g CS 6.7 g Cl 70.906 g Cl 3 mole Cl 1 mole CS =. g CS 13. Titanium(IV) chloride is obtained from titanium(iv) oxide by the following process: 3 TiO (s) + C(s) + 6 Cl (g) 3 TiCl (g) + CO (g) + CO(g) A vessel contains.15 g TiO, 5.67 g C and 6.78 g Cl. How many grams of titanium(iv) chloride can be produced? 1 mole TiO 3 mole TiCl 189.679 g TiCl.15 g TiO 79.865 g TiO 3 mole TiO 1 mole TiCl = 9.86 g TiCl 1 mole C 3 mole TiCl 189.679 g TiCl 5.67 g C 1.011 g C mole C 1 mole TiCl = 67. g TiCl 1 mole Cl 3 mole TiCl 189.679 g TiCl 6.78 g Cl 70.906 g Cl 6 mole Cl 1 mole TiCl = 9.07 g TiCl