Question & Answer: Calculate the empirica lormula , , Calculate the mass percentage of nitrogen in the following:…..

Q4.

MW of N = 14 g/mol

Mw of Ca(NO3)2= 164.088 g/mol

then % of N = 2*14/(164.088)*100% = 17.06401 %

b)

Note that % of Nitrogen should not change with respect of total amount of mass, since proportios are kept

i.e. 1 mol Ca(NO3)2 contains always 2 mol of N

c)similar as b

Special note for Question 4: the value of “a” you calculated is correct, all other values are incorrect, since the raito is kept, there is no change in % of N in Ca(NO3)2

Question 5.

a)

2.89 g of Gold = x mol

1 mol of Gold = 196.96657 g

x mol = 2.89/(196.96657) = 0.014672 mol of Gold

1 mol = 6.022*10^23 (avogrados number)

0.014672 mol = (0.014672)(6.022*10^23) =8.835*10^21 atoms of Gold (your answer is correct)

b)

2 lb = 2*454 g = 908 g

mol of Mg = mass/MW = 908/24.305 = 37.358 mol of Mg

1 mol = 6.022*10^23

37.358 mol = (37.358)(6.022*10^23) = 2.2496*10^25 atoms… your answer is correct

c)

4.3 mol of Tungsten = 4.3*6.022*10^23 = 2.58*10^24 atoms of tungsten, also correct

d)

4.15 mg = 4.15*10^3 g of HgI2

mol = masS/MW = (4.15*10^-3)/(454.4) = 0.0000091329 mol of HgI2

1 mol of HgI2 = 3 mol of atoms, i.e. Hg, I and I

0.0000091329 mol = 0.0000091329*3 = 0.0000273987 mol of atoms

1 mol = 6.022*10^23

0.0000273987 mol = (0.0000273987)(6.022*10^23) = 1.6499*10^19 atoms, your answer is incorrect

your answer lacks the multiplication of: 1 mol of HgI2 = 3 mol of atoms

e)

Cu(NO3) total atoms = 1+1+3*1 = 5 atoms

mol of CuNO3 = mass/MW = (4.25)/(125.5509) = 0.03385 mol of CuNO3

mol ofatoms = 5*0.03385 = 0.16925 mol of atoms

now,

1 mol = 6.022*10^”3

0.16925 mol = (0.16925)(6.022*10^23) = 1.019*10^23 atoms present, once again you must multiply by 5 since 1 mol of CuNO3 contains 5 atoms