Review Material For Exam I

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1 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 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 in (1 in) 1 cm 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

2 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 Fe 3+ 6 Ne O Au Br P Os Hg Pb 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 g Br 1 mole Br 1 molecule Br =.1 x 10 3 atoms Br

3 c. number of molecules in 3.0 g of NH 3 1 mole NH 6.0 x 10 molecules NH 3.0 g NH g NH 1 mole NH 3 = 1.5 x 10 molecules NH 3 d. number of molecules in g CCl 1 mole CCl 6.0 x 10 molecules CCl g CCl 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 ) g Cr (SO ) 1 mole Cr (SO ) = 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 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/ g H 1 mole C 6.7 g C x.30 mole C/ g C 1 mole O 71.1 g O x.0 mole O/ g O empirical formula = HCO FW = g/mole g/mole ; molecular formula = x empirical formula = H C O

4 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/ x g C = 1 mole H 6.90 g H x 6.85 mole H/ x g H = 1 mole O 3.8 g O x.7377 mole O/ x g O = empirical formula = C 3 H 5 O FW = g/mole 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 O 8 CO + 10 H O 10 mole H O 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 CO How many grams of calcium phosphate are required to give 5.00 g of phosphorous? 1 mole P mole Ca (PO ) g Ca (PO ) g P g P 1 mole P 1 mole Ca 3(PO ) = 5.0 g Ca 3 (PO )

5 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 g CS 6.7 g Cl 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 g TiCl.15 g TiO g TiO 3 mole TiO 1 mole TiCl = 9.86 g TiCl 1 mole C 3 mole TiCl g TiCl 5.67 g C g C mole C 1 mole TiCl = 67. g TiCl 1 mole Cl 3 mole TiCl g TiCl 6.78 g Cl g Cl 6 mole Cl 1 mole TiCl = 9.07 g TiCl

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