Question

A ball bearing similar to problem 1.1 has the following composition of alloying elements (once again tne remainder is iron) and Has a weight of 3.756g, If we removed 12 sextillion atoms from the ball bearing new much would the remaining chromium in the modified ball bearing weigh? The answer should be in grams and have 3 decimals of accuracy.

Answer 1

Calculate weight % of iron as follows: iron 00 100-sum of weight % (S + Si + P + Mn + C + Mo + Cr) = 100-(0.043 0.220 + 0.031 + 0.546 + 0.300 + 0.238 + 0.900) 100-2.278 97.72% Calculate weight and moles of each element respectively in 3.756 g of ball as follows: 1. For Sulfur, Weight weight %x weight of ball 100 0.043 x 3.756 g 100 = 1.6×10-3 g Mole - weight of element molar mass of element 1.6x10-3 g 32.065 g mol 4.99 x10-mol

2. For Silicon, -weight % weight of ball 100 Weight- 0.220 x 3.756 g 100 = 8.2×10-3 g weight of element molar mass of element Mole = 8.2x10 g 28.085 g mol 2.92 × 10-4 mol 3. For Phosphorous, Weight-weight weight of ball 100 0.031x 3.756 g 100 1.16x10g weight of element molar mass of element 1.16x10-1 g 30.97 g mol Mole- -3.7x105 mol

4. For Manganese, Weight = weight % weight of ball 0.546 x 3.756 g 2.05x102g weight of element 100 100 Mole- molar mass of element 2.056x10g 54.938 g mol1 = 3.74 × 10-4 mol 5. For Carbon, weight %× weight of ball Weight = 100 0.300 x 3.756 g 100 -1.126x102 g weight of element molar mass of element Mole = 1.126x102 g 12.0107 g mol1 = 9.37 × 10-4 mol

6. For Molybdenum, Weight-weight weight of ball 100 %x 0.238 x 3.756 g 100 = 8.9×10-3 g weight of element molar mass of element 8.9x10-3 g 95.94 g mol = 9.27×10-5 mol 7. For Chromium, Weight-weight weight of ball 100 0.900 x 3.756 g 100 = 3.38×10-2 g Mole weight of element 3.38x10 g 51.996 g mol 6.5×10-4 mol molar mass of element

7. For Iron, weight %× weight of ball 9112 × 3756 g 3.67 g Weight = 100 100 weight of element Mole = molar mass of element 3.67 g 55.845 g mol = 6.5×10-2 mol Find the smallest integer ratio of each element by dividing their moles by the smallest moles obtained (sulfur) as follows: 4.99x10- mol 4.99 x 10 mol Sulfur 2.92 x10-4 mol 4.99×10-5 mol Silicon = 5.8

3.7x10- mol 4.99 x 10- mol -0.74 3.74x10 mol Manganese = 4.99× 10-5 mol 9.37x10 mol 4.99×10-5 mol Carbon -18.7 = 19 9.27x10- mol 4.99x10 mol Molybdenum- = 1.85 6.5×10-4 mol 4.99 x10- mol Chromium- = 13.02 - 13 Iron 6.5x102 mol 4.99x10-5 mol 1302.6 1 302

Let x be the number of atoms in each element removed. Since 12 sextillion atoms are removed, therefore lx + 6x + lx +7x+19x + 2x + 13x + 1302x = 12x1021 atoms 1351x=12×1021 atoms 21 12x10 atoms 1351 -8.8×1018 atoms Therefore, number of Chromium atoms removed is: Cr 13x 8.8x1018 atoms -114.4x1018 atom Convert into moles as follows: Moles of Cr = 114.4x101s atoms 6.023×1023 atoms mol-1 1 .89 × 1 0-4 mol Hence, remaining number of moles of Chromium is: = 6.5x10-4 mol-1.89x10-4 mol = 4.61x 10-4 mol Convert into mass: 1 mol 51.996 g mol Cr 4.61 × 10-4 mol 51.996 g mol-ı × 4.61 × 10-4 mol 2.397x102 g = 0.024 g