Question

(a) mass, in grams, of 0.105 mole of sucrose (C12H22O11)

(b) moles of Zn(NO3)2 in 143.50 g of this substance

(c) number of N atoms in 0.410 mol NH3

Answer 1

a) If you have a moles and a compound formula presented, you might as well multiply moles by the atomic mass of the compound formula. Do you know how to derive a mass from a chemical formula by itself? If you know the molecular weight of sucrose (C12H22O11), then you're good to go
0.105 moles sucrose times grams of sucrose in molecular weight)= mass of grams of sucrose

But FIRST, find the molecular weight of grams C12H22O11in 1 mole.

b) Here, you are asked to calculate the moles of Zn(NO3)2 in 143.50 g of the same substance of Zinc Nitrate. To solve this, we need to multiply 143.50 g Zn(NO3)2 by (1 mol Zn(NO3)2) divded by (189.39g/mol)
-- 189.39g/mol is the calulcated mass for Zinc Nitrate
--The unit cancel out, so we have the units in mols 143.50 g Zn(NO3)2 x1 mol Zn(NO3)2189.39 g of Zn(NO3)2

143.50 g Zn(NO3)2 x1 mol Zn(NO3)2189.39= g of Zn(NO3)2


c) Are you familiar with Avogadro's number? Avogadro's are used to convert moles to molecules. 6.022 x 10^23 is Avogadro's number

So, we have number of N atoms in 0.410 mol NH3 How many atoms we have in NH3? There's 1 N and 3 H Atoms. Ok, now if 1 mole consists of 6.022 x 10^23 atoms,that means
6.022x10^23 molecules divide by 1 mol (0.410 mol NH3) = Nitrogen Atoms
and 6.022x10^23 molecules/3 mol (0.410 mol NH3)= Hydrogen atoms
---but here, u are trying to solve for N which is Nitrogen :)