CHAPTER 7 Gameday Review.

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Presentation transcript:

CHAPTER 7 Gameday Review

MOLE CONVERSIONS Determine the number of atoms present in 1.55 mol of sodium 9.33 x 1023 atoms Na

MOLE CONVERSIONS Determine the mass in grams of 0.750 mol of francium. 167 g Fr

MOLE CONVERSIONS A sample of lead has a mass of 150.0 g. How many atoms does the sample contain? 4.360 x 1023 atoms Pb

MOLE CONVERSIONS Convert 2.65 x 1025 atoms of fluorine to grams of fluorine. 8.36 x 102 g F

MOLAR MASS Calculate the molar mass of phenol (C6H5OH) 94.12 g / mol

Molar Mass Sodium Sulfate Na2SO4 = 142.05 g/mol

AVERAGE ATOMIC MASS Calculate the average atomic mass for the isotopes below to 5 decimal places: mass number exact weight percent abundance 12 12.000000 98.90 13 13.003355 1.10 12.01104 amu

Average Atomic Mass Calculate the average atomic mass for the isotope below and take it out 5 decimal place: mass number exact weight percent abundance 14 14.003074 99.63 15 0.37 14.98189 amu

EMPIRICAL FORMULA What is the empirical formula for a compound that is 26.56% Potassium, 35.41 % Chromium, and 38.03% oxygen? K2Cr2O7

EMPIRICAL FORMULA What is the empirical formula for a compound that is 29.44% calcium, 23.55% sulfur, and 47.01% oxygen? This compound is an ingredient in plaster. CaSO4

MOLECULAR FORMULA A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound? C10H7O2 – empirical formula C20H14O4 – molecular formula

MOLECULAR FORMULA Ribose is a type of sugar consisting of 40% carbon, 6.7% hydrogen, and 53.3 % oxygen. A mass spectrometer was used to determine the molar mass of ribose to be 150 g/mol. What is the molecular formula for ribose ? CH2O – empirical formula C5H10O5 – molecular formula

% COMPOsition Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. mass % Na = 22.99 g / 84.01 g x 100 = 27.36 % mass % H = 1.01 g / 84.01 g x 100 = 1.20 % mass % C = 12.01 g / 84.01 g x 100 = 14.30 % mass % O = 48.00 g / 84.01 g x 100 = 57.14 %

CONGRATULATIONS!!!!! STUDY! STUDY! STUDY! STUDY! Molar Mass Problems Mole Conversions Percent Composition Average atomic mass Molecular and Empirical formulas