Question

I need help finding the experimental formula unit, theoretical formula unit and writing balanced chemical equations

CHEM 1211 Lah Mamal Revised 5eis Mole of oxygen in compound o cos Ratio: mole of magnesium : mole of oxygen (Show in this order, mol of Mg, colon Cl mole of O) o.ooi니 : 0 , 00631 Ratio: mole of magnesium : mole of oxygen expressed as a simple, whole number ratio (Show in this order, mol of Mg. colon C, mole of O) mol Mg 0:3 Experimental formula unit of the compound (5 pts.) (based on experimentally determined ratio) Theoretical formula unit of compound (based on ion charges) Write balanced chemical equations for the three chemical reactions that occur inside the crucible. Be sure to use correct theoretical formulas and include the physical state of each substance as (s). (D. (g), or (aq) for both reactants and products. Look for hints to products in the Introduction (first page of this assignment). ((4 pts. each) Formation of the magnesium-oxygen compound Formation of the magnesium-nitrogen compound Reaction of the magnesium-nitrogen compound with water In complete sentences, record observations during the experiment, including sources of experimental errors. You may use the back of this page, if needed. (10 pts)

Answer 1

According to your experimental data

mass of Mg = 25.626-25.591 = 0.035 g

and mass of Magnesium oxygen compound = 25.631 - 25.591 = 0.04 g

mass of oxygen = mass of Magnesium oxygen compound - mass of magnesium in it

= 0.04 - 0.035 = 0.005 g

(1)

moles of oxygen in compound =

According to your experimental data mass of Mg = 25.626 - 25.591 = 0.035 g and mass of Magnesium oxygen compound = 25.631 - 25.591 = 0.04 g mass of oxygen = mass of Magnesium oxygen compound - mass of magnesium in it = 0.04 - 0.035 = 0.005 g moles of oxygen in compound = 0.005/16 = 3.125 times 10^-4 moles moles of magnesium in compound = 0.035/24 = 1.458 times 10^-3 moles moles ratio of Mg: O is 1.458 times 10^-3: 3.125 times 10^-4 write it in smallest whole number ratio 145: 31 so the experimental formula of the compound is Mg_145 O_31 theoretical formula based on change. Suppose formula is Mg_x O_y \r\n now apply the charge balance: Mg usually shows + 2 oxidation state and O shows -2 oxidation state so charge balance = > + 2 times x + (-2)y = 0 x/y = 1 this means the ration of Mg: O is 1: 1 so formula is MgO. Balanced equation. Formation

moles of magnesium in compound = $\frac{0.035}{24} = 1.458 \times 10^{-3} \; \; moles$

moles ratio of Mg : O

is 1.458 $\times$ 10^-3 : 3.125 $\times$ 10^-4

write it in smallest whole number ratio

145 : 31

so the experimental formula of the compound is

Mg₁₄₅O₃₁

theoritical forumula based on charge,

suppose formula is Mg_xO_y

now apply the charge balance ; Mg usually shows +2 oxidation state and O shows -2 oxidation state so charge balance $\Rightarrow$ $+2\times x + (-2) y = 0$

$\frac{x}{y}=1$

this means the ration of Mg : O

is 1 : 1

so formula is MgO.

Balanced equation.

1. formation of magnesium oxygen compound

2 Mg (s) + O₂ (g)   $\rightarrow$ 2 MgO (s)

how we decided product i have already shows above Mg shows +2 oxidation state and O shows -2.

2. formation of magnesium -Nitrogen compound

3 Mg (s) + N₂ (g) $\rightarrow$ Mg₃N₂ (s)

Mg shows +2 oxidation state and Nitrogen usually shows -3 oxidation state. so ratio

assume formula of product $Mg_xN_y$ , apply charge balance

$\Rightarrow 2 \times x + (-3) y = 0$

   $\Rightarrow \; \; \; \; \; \; \; \frac{x}{y} = \frac{3}{2}$

so ratio Mg : N = 3:2

formula = Mg₃N₂

3. reaction of magnesium nitrogen compound with water

it's already mentioned in question that the product is desired magnesium oxygen compound (MgO) and ammonia gas (NH₃)

Mg₃N₂ (s) + 3 H₂O (l) $\rightarrow$ 3 MgO (s) + 2 NH₃ (g)