are my answers correct?

Calculate the empirica lormula , , Calculate the mass percentage of nitrogen in the following: / 4. . a. 1.00 mol Ca(NO3)2-70/ b. c. 0.67 mol Ca(NO3)27 h,UY% 2.71 mol Ca(NO3)2 6.2lo 5. Using the average atomic masses. Calculate the number of atoms present in each of the following samples 10っs 2.0 lb of magnesium a2 c. 4.30 moles of tungsten 2.6×12 d. 4.15 mg mercury (II) iodide 5 ,グメレ e. 4.25 g copper (I) nitrate b.

## Expert Answer

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