Problem Solving 1

Slide 1: How to Solve Problems and Consider Significant Figures Special Issue

Slide 2: Every problem has its own “anatomy”. A problem consists of tale and data. Tale may be boring and funny. Data may be necessary and unnecessary.

Slide 3: The tale and the data: Silver(Ag: Z = 47) has 46 known isotopes, but only two occur naturally, 107Ag and 109Ag. Given the following mass spectrometric data, calculate the atomic mass of silver: 107 Ag: atomic mass 106.90509, abundance 51.84 109 Ag: atomic mass 108.90476, abundance 48.16

Slide 4: Let us handle a problem step by step: Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral.

Slide 5: 1.Read the problem carefully: Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral.

Slide 6: 2. Flesh out the given data: Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral. m1(Zn)=327.1 g ; m1(S)=160.4 g ; m(ore)=541 kg

Slide 7: 3. Understand what is required: Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral. m2(Zn)= ? ; m2(S)= ?

Slide 8: 4.Speculate starting from the question posed: To find mass of zinc and sulfur in a given mass of their compound, we must know their mass fractions (or mass percents) in this compound. To find the mass fraction of zinc, we should divide mass of zinc in the sample by the mass of this sample. The same about sulfur. To find the mass of the sample, we have to add masses of zinc and sulfur in this sample.

Slide 9: 5. Solve (slide 1) starting from the end of the speculation: Find the mass of the ore sample: 327.1g + 160.4g = 487.5g Find the mass fraction of zinc: 327.1g/487.5g = 0.67097436 Consider the number of significant figures: 3. Round the value of zinc mass fraction: 0.67097436 ≈ 0.671

Slide 10: 5. Solve (slide 2) starting from the end of the speculation: Find the mass fraction of sulfur: 160.4g/487.5g = 0.32902564 Consider the number of significant figures: 3. Round the value of sulfur mass fraction: 0.32902564 ≈ 0.329 or just 1.000 - 0.671 = 0.329 - much faster!

Slide 11: 5. Solve (slide 3) starting from the end of the speculation: Find the mass of zinc in 541 kg of the mineral: 0.671 x 541 kg = 363.011 kg Consider the number of significant figures: 3. Round the mass of zinc: 363.011kg≈363 kg Convert to grams: 363kg=363000g=3.63x105 g Finalize for WebAssign: 3.63e5

Slide 12: 5. Solve (slide 4) starting from the end of the speculation: Find the mass of sulfur in 541 kg of the mineral: 0.329x541kg=177.989kg Consider the number of significant figures: 3. Round the mass of sulfur: 177.989kg≈178kg Or just 541 kg - 363 kg = 178 kg - much faster! Convert to grams:178kg=178000g=1.78x105g Finalize for WebAssign: 3.63e5

Slide 13: Let us handle another problem step by step: Gallium has two naturally occurring isotopes, 69Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium.

Slide 14: 1.Read the problem carefully: Gallium has two naturally occurring isotopes, 69Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium.

Slide 15: 3. Understand what is required: Gallium has two naturally occurring isotopes, 69Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium.

Slide 16: 4.Speculate starting from the question posed: To find the atomic mass of gallium, we must add the portions contributed by each isotope of gallium. To find a portion contributed, we have to multiply an isotope’s mass number by its fractional abundance.

Slide 17: 5. Solve (slide 1) starting from the end of the speculation: Find the portion contributed by 69Ga: 68.9256x60.11% / 100%=41.431178 or 68.9256x0.6011=41.431178 Consider the number of significant figures: 4. Round the value of 69Ga portion: 41.431178 ≈ 41.43

Slide 18: 5. Solve (slide 2) starting from the end of the speculation: Find the portion contributed by 71Ga: 70.9247x39.89% / 100%=28.291863 or 70.9247x0.3989=28.291863 Consider the number of significant figures: 4. Round the value of 71Ga portion: 28.291863 ≈ 28.29

Slide 19: 5. Solve (slide 3) starting from the end of the speculation: Add the portions contributed by 69Ga and 71Ga: 41.43+28.29=69.72 . It is atomic mass of gallium.

Slide 20: THE END