Chapter 3 Lecture

(make sure that percents add up to 100 or approximately 100) 1. Gallium consists of two isotopes of masses 68.95 amu and 70.95 amu with abundances of 60.16% and 39.84%, respectively. What is the atomic mass of gallium? Atomic mass = (mass I1)(% I1) + (mass I2)(% I2) + (continued) (make sure that percents add up to 100 or approximately 100) = (68.95 amu)(0.6016) + (70.95 amu)(0.3984) = 69.75 amu

2. Magnesium exists as three isotopes in nature 2. Magnesium exists as three isotopes in nature. One isotope (25Mg) has a mass of 24.99 amu and a relative abundance of 10.13%. The other two isotopes have masses of 23.99 amu (24Mg) and 25.98 amu (26Mg). What are their relative abundances? 24.31 = (24.99)(0.1013) + (23.99)(% I2) + (25.98)(% I3) 24.31 = (24.99)(0.1013) + (23.99)(x) + (25.98y) 0.1013 + x + y = 1 (all % abundances add up to 100%) x = 0.8987 – y 24.31 = (24.99)(0.1013) + (23.99)(0.8987 - y) + (25.98y) y = 10.99%, x = 78.88 %

3. Calculate the molar masses of the following compounds: a 3. Calculate the molar masses of the following compounds: a. Zr(SeO3)2 b. NH4OH c. Ca2Fe(CN)6.12H20 40.08(2) + 55.85 + 12.01(6) + 14.01(6) + 1.01(24) + 16.00(12) 508.37 g/mole 91.22 + 78.96(2) + 16.00(6) 345.14 g/mole 14.01 + 1.01(5) + 16.00 35.06 g/mole

4. What are the mass percents of iron and oxygen in Fe2O3? Iron: (2 x 55.85 g/mole) x 100= 69.94 % (159.7 g/mole) Oxygen: (3 x 16.00 g/mole) x 100 = 30.06 %

Use the strategy we just learned in #3! Two ways of solving the same problem 5. How many grams of iron can be extracted from 567 grams of iron(III) oxide? 567 g Fe2O3 ( 1 mole Fe2O3 ) ( 2 moles Fe) (55.85 g Fe) = (159.7 g Fe2O3) (1 mole Fe2O3) (1 mole Fe) OR Use the strategy we just learned in #3! mass of compound x % iron in the = mass iron in in sample compound the sample (567 g Fe2O3)( 0.6994) = 397 g Fe

6. A substance contains 23. 0 g sodium, 27 6. A substance contains 23.0 g sodium, 27.0 g aluminum, and 114 g fluorine. How many grams of sodium are there in a 120 gram sample of the substance? Total mass: 164 g 23.0 g Na = 0.140 or 14.0% 164g (0.140)(120g) = 17 g Na

7. A 25. 0 gram sample of a compound contains 6. 64 grams potassium, 8 7. A 25.0 gram sample of a compound contains 6.64 grams potassium, 8.84 g chromium, 9.52g oxygen. Find the empirical formula of this compound. Step 1: Convert each mass to moles 0.170 mole K 0.170 mole Cr 0.595 mole O 0.170 0.170 0.170 Step 2. Get ratio of moles (by dividing each # moles by smallest # moles) Mole ratio: 1 mole K: 1 mole Cr: 3.50 moles O Step 3: If the numbers in the ratio are not integers, double, triple, etc each number until they are all integers….you may want to round slightly (ex. 2.98 can be 3 but 2.49 should be doubled to 4.98, then rounded to 5) New ratio: 2 : 2 : 7 K2Cr2O7, potassium dichromate

8. Phenol is a compound which contains 76. 57% carbon, 6 8. Phenol is a compound which contains 76.57% carbon, 6.43% hydrogen, and 17.0% oxygen. What is the empirical formula of phenol? If you are given percents of elements instead of actual masses, turn each percent into a gram amount out of a 100g sample Step 1: Convert each mass to moles 76.57g C/12.01 g C 6.43 g H/1.01g H 17.0 g O/16.00 g O 6.376 mole C 6.366 mole H 1.06 mole O 1.06 1.06 1.06 Step 2. Get ratio of moles (by dividing each # moles by smallest # moles) Mole ratio: 6 mole C: 6 mole H: 1 moles O C6H6O = phenol!

molecular formula = multiple of the empirical formula 9. The empirical formula of styrene is CH; its formula weight (‘molar mass’) is 104.1 g/mole. What is the molecular formula of styrene? molecular formula = multiple of the empirical formula * molar mass of the compound is needed Molar mass styrene = 104.1 g/mole Molar mass emp.form. = 13.02 g/mole = 7.995391… = 8 molecular formula = C8H8

10. A 5. 00 gram sample of an acid contains 2. 00 g carbon, 0 10. A 5.00 gram sample of an acid contains 2.00 g carbon, 0.336 g hydrogen, and 2.66 g oxygen. Find the molecular formula of this acid and give the proper name if its molar mass = 60.06 g/mole Try this yourself: answer: acetic acid

# molecules compound in sample Mass compound in sample The Molar Highway Finding Various Quantities using The Mole mass of element or ion in compound # atoms or ions in sample moles of element or ion in the compound Need: # of that atom or ion in compound Need: atomic mass of element or ion Need: # of that atom or ion in compound # molecules compound in sample Mass compound in sample 1 mole of a compound Need: molar mass of compound Need: 1 mole = 6.02 x 1023 particles

11. A sample of 3.9 x 1019 strontium chloride molecules contains how many milligrams of chlorine? 3.9 x 1019 molec. SrCl2 (2 atoms Cl ) (1 mole Cl) ( 35.45 g Cl) ( 1000 mg Cl) = (1 molec SrCl2) (6.02 x 1023 atoms Cl) (1 mole Cl) ( 1 g Cl) 4.6 mg chlorine

12a. What mass of antimony(III) iodide can form from 1.20g Sb ? 2Sb(s) + 3I2(s) → 2SbI3(s 12a. What mass of antimony(III) iodide can form from 1.20g Sb ? 2 1.20 g Sb ( mol Sb ) 1 ( mol SbI3) ( g SbI3 ) 502.46 1 ( g Sb ) 121.76 ( mol Sb) 2 ( mol SbI3 ) = 4.94 g SbI3 Wait! We know how much Sb we have. But what if there is just a tiny, tiny amount of iodine available? Hmmmm…

wait but 1.20 g Sb predicts 4.94 g SbI3 2Sb(s) + 3I2(s) → 2SbI3(s) 12b. Determine the theoretical yield (mass in grams) of antimony(III) iodide formed when 1.20g Sb and 2.40 g I2 are mixed. 2.40 g I2 ( mol I2 ) 1 ( mol SbI3) 2 ( g SbI3 ) 502.46 ( g I2 ) 253.8 ( mol I2) 3 ( mol SbI3 ) 1 = 3.17 g SbI3 wait but 1.20 g Sb predicts 4.94 g SbI3 5. Do another calculation with the other amount given 6. Choose the lower amount…the higher amount is not possible. The theoretical yield is calculated based upon the limiting reactant. 3.17 g SbI3. I2 is the limiting reactant (even though there is a greater amount of it) and is completely used up. Some Sb will be leftover.

12c. If the experiment actually produces 3 12c. If the experiment actually produces 3.00g SbI3, calculate the percent yield of the experiment. percent yield = actual or experiment yield x 100 theoretical yield percent yield = 3.00 g x 100 3.17 g 94.6 % yield