Find the empirical formula of each of the following compounds. given mass or for each element in a sample of the compound 3,611 g ca; 6.389 g c1

CaCl₂  

Step-by-step explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ca to Cl.

Data:

Mass of Ca = 3.611 g

Mass of Cl = 6.389 g

Calculations

Step 1. Calculate the moles of each element

Moles of Ca = 3.611 g Ca × (1 mol Ca/(40.08 g Ca)= 0.090 10 mol Ca

Moles of Cl = 6.389 g Cl  

Step 2. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

Ca:Cl = 0.090 10:0.1802 = 1:2.000

Step 3. Round the molar ratios to the nearest integer

Ca:Cl = 1:2.000 ≈ 1:2

Step 4: Write the empirical formula

EF = CaCl₂

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

that tool would be a compass