# 1. The composition of a compou

1. The composition of a compound is 34.6% carbon, 3.9% hydrogen,and 61.5% oxygen . What is the empirical formula of thecompound?

2. A mixture consisting of 15.0 g of iron(II) sulfate and 15.0 gof sodium phosphate is reacted in the following reaction:

3 F e S O 4 ( a q ) + 2 N a 3 P O 4 ( a q ) ⟶ F e 3 ( P O 4 ) 2( s ) + 3 N a 2 S O 4 ( a q )

Formula weight of FeSO4 = 151.91 g/mol

Formula weight of Na3PO4 = 163.94g/mol

Formula weight of Na2SO4 = 119.05g/mol

• What is the maximum number of grams of sodium sulfate that canbe obtained?
• Calculate the grams remaining of the excess reactant.

1)

Let mass of compound be 100gm

Mass of carbon = 34.6 gm

Mole of carbon = 34.6 / 12 = 2.883

Mass of H = 3.9 gm

Mole of H = 3.9 / 1.00784 = 3.87

Mass of O = 61.5 gm

Mole of O = 61.5 / 16 = 3.84

C2.88H3.87O3.84

Emperical formula :- C3H4O4

2)

Mole of FeSO4 = mass / molar mass

= 15 / 151.91 = 0.0987

Mole of Na3PO4 = mass / molar mass

= 15 / 163.94 = 0.0915

(Mole/coefficient)FeSo4 is less than that of Na3PO4

So, FeSO4 is limiting reagent.

(Mole/coefficient)FeSO4 = (Mole/coefficient)Na2SO4

Mole of FeSO4 / 3 = mole of Na2SO4 / 3

Mole of Na2SO4 = mole of FeSO4 = 0.0987

Mass of Na2SO4 formed = 0.0987 * 119.05

= 11.75 gm

Remaining mole of excess Na3PO4

= 0.0915 – 2* 0.0987/3

= 0.0915 – 0.0658 mole

= 0.0257 mole

Mass of Na3PO4 remains = remaining mole * molar mass

= 0.0257 * 163.94

= 4.213 gm

##### "Our Prices Start at \$11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"

Pages (275 words)
Standard price: \$0.00
Client Reviews
4.9
Sitejabber
4.6
Trustpilot
4.8
Our Guarantees
100% Confidentiality
Information about customers is confidential and never disclosed to third parties.
Original Writing
We complete all papers from scratch. You can get a plagiarism report.
Timely Delivery
No missed deadlines – 97% of assignments are completed in time.
Money Back