Question

9:41 A chromoly (a 41xx series) steel ball bearing weighs 4.224 grams. It contains the following weight fraction of elements added to iron shown in the table below. (note: all of the elements plus the iron should add up to 100 wt%). Determine how many atoms are in the ball bearing. The answer should be in units of moles with three decimals of accuracy Component weight % Sulfur Silicon Phosphorus 0.024 Manganese 0.546 Carbon Molybdenum 10.205 Chromium .925 0.043 16 0.319

Answer 1

Total weight percent without iron=(0.043+0.16+0.024+0.546+0.319+0.205+0.925)=2.22

Weight percent of iron=(100-2.22)=97.778

Amount of S=(0.043 × 4.224)÷100= 0.0018 g=0.0018÷32=0.00056 moles

Amount of silicon=(0.16×4.224)÷100=0.0067 g=0.0067÷28.08=0.00024 moles

Amount of phosphorus=(0.024×4.224)÷100=0.001 g=0.001÷30.973=0.00032 moles

Amount of manganese=(0.546×4.224)÷100=0.023 g=0.023÷54.94=0.00042 moles

Amount of carbon=(0.319×4.224)÷100=0.0134 g=0.0134÷12=0.0011 moles

Amount of molybdenum=(0.205×4.224)÷100=0.0086g=0.0086÷95.94=0.000902 moles

Amount of chromium=(0.925×4.224)÷100=0.039g=0.039÷51.996=0.00075 moles

Amount of iron=(97.778×4.224)÷100=4.13 g=4.13÷55.845=0.074 moles

Total moles=0.00056+0.00024+0.00032+0.00042+0.0011+0.000902+0.00075+0.074=0.0783 moles

1mole=6.023E23 atoms/mole

0.0783 moles=4.71E22 atoms/mole