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Practice Mole Problems

Aluminum is a metal with high strength-to-mass ratio and a high resistance to corrosion; thus it is often used for structural purposes.


 Compute both the number of moles of atoms and the number of atoms in a 10.0g sample of aluminum.Solution:1 mole of Al will contain 26.981539 g of AlSo, how many moles of Al will contain 10 grams of Al?If 'x' is the number of moles in 1g of Al, x = (1 mole) × (10 g) / (26.981539 g)x = 0.371 moles of AlHaving found this, we now need to move on to how many atoms of Al are present in 0.371 moles of Al.We know that 1 mole of Al contains 6.023 × 10 23 atoms of AlSo, in 0.371 moles of AL, 'y' atoms of Al are present.y = (0.371 moles) (6.023 × 10 23atoms) /(1 mole)y = 2.23 × 10 23 atoms of AlThe answer therefore is 2.23 × 10 23 atoms of Al. 


 Cobalt is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00 × 10 20 atoms and the mass of the sample.Solution:1 mole of cobalt will contain Avagadro number of atoms, that is 6.023 × 10 23 atoms.So, 5.00 × 10 20 atoms will be present in 'a' number of moles.a = (5.00 × 10 20 atoms)(1 mole ) / (6.023 × 10 23 atoms)a = 8.30 × 10 -4 molesNow to the second part of the question. 


The molar mass of cobalt is 58.93320 g. Or in other words, 1 mole of cobalt will contain 58.93320 g of Co. The given sample contains 8.30 × 10 -4 moles of Co. We need to find the mass of 8.30 × 10 -4 moles of Co.Let the value be 'b'.b = (8.30 × 10 -4 moles) (58.93320 g) /(1 mole)b = 0.0489g of Coor b = 4.89 × 10 -2 g of CoA few more mole problems for practiceWhat number of Fe atoms and what amount (moles) of Fe atoms are in 500.0g of iron?Solution:1 mole of iron (Fe) contains 55.847 g of Fe.Therefore, in 'p' moles of Fe, 500 g of Fe are presentp = (500 g) (1 mole) / (55.847 g)p = 8.953 moles of FeA diamond contains 5.0 × 10 21 atoms of carbon. What amount (moles) of carbon are in this diamond?Solution:1 mole of C contains 6.023 × 10 23 atoms.So, 5.0 × 10 21 atoms of carbon will be present in 'q' moles of C.q = (5.0 × 10 21 atoms) (1 mole) / (6.023 × 10 23 atoms)q = 8.30 × 10 -3 moles of C