Question

A. How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 342.6 mL of a 0.783 M LiNO3 solution? B. What mass of H3PO4 (98.0 g/mol) is present in 36.2 L of a 0.0827 M solution of H3PO4? C. The formula mass of zinc acetate trihydrate, Zn(CH3COO)2 • 3H2O, is ? D. A 2.39-g sample of an oxide of chromium contains 1.48 g of chromium. Calculate the simplest formula for the compound E. The analysis of an organic compound showed that it contained 1.386 mol of C, 0.0660 mol of H, 0.924 mol of O, and 0.462 mol of N. How many nitrogen atoms are there in the empirical formula for this compound? Thanks

Answer 1

How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 342.6 mL of a 0.783 M LiNO3 solution?

moles of LiNO3 in 342.6 ml fof 0.783M= Molarity* Volume(L) =0.783*0.3426=0.27

Moles = mass/molar mass, mass of LiNO3 = moles* molar mass =0.27*68.9=18.6 gm

2.

What mass of H3PO4 (98.0 g/mol) is present in 36.2 L of a 0.0827 M solution of H3PO4

Moles in H3PO4= 0.0827*36.2= 2.994, mass of H3PO4= 2.994*98=293.4 gm

3. Zn(CH3COO)2 • 3H2O

4. A 2.39-g sample of an oxide of chromium contains 1.48 g of chromium. Calculate the simplest formula for the compound

mass of chromium = 1.48 gm , mass of O = 2.39-1.48=0.91

atomic weights : Cr= 52 and O=16

Moles = mass/molar mass

Moles : Cr= 1.48/52= 0.0284 and O=0.91/16=0.057

Cr: O =0.0284 : 0.057 = 1:2

So the formula id CrO2

4.he analysis of an organic compound showed that it contained 1.386 mol of C, 0.0660 mol of H, 0.924 mol of O, and 0.462 mol of N. How many nitrogen atoms are there in the empirical formula for this compound? Thanks

Molar ratio of C: H:O :N = 1.386 : 0.066 :0.924 : 0.462

dividing by 0.066 the ratio becomes 1.386/0.066 :0.066/0.066 : 0.924/0.066 :0.462/0.066 = 21:1: 14: 7

So the empirical formula is C21HO14N7. So there are 7 atoms of Nitrogen