A sample of a compound containing C, O, and silver (Ag) weighed 1.372 g. On analysis, it was found to contain 0.288 g O and 0.974 g Ag. The molar mass of the compound was determined to be 308.8 g/mol. What is the molecular formula of the compound?

The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.

Explanation:

This is an easy excersise, we relate the moles of the elements in the compound with the molar mass of the total compound.

The sample weighs 1.372 g, with 0.288 g of O, 0.974 g of Ag and the rest, C

Therefore, 1.372g - 0.288 g - 0.974g = 0.112 g are C

We convert the mass to moles

0.288 g / 16g/mol = 0.018 moles of O

0.974 g / 107.87 g/mol = 0.00903 moles of Ag

0.112 g / 12 g/mol = 0.00933 moles of C

Now, the use of the molar mass of the compound. Let's prepare some rules of three:

1.372 g of compound contain 0.018 moles of O, 0.00903 moles of Ag and 0.00933 moles of C

Then, 308.8 g of compound (a mol) must contain:

(308.8 . 0.018) / 1.372 = 4 moles of O

(308.8 . 0.00903) / 1.372 = 2 moles of Ag

(308.8 . 0.00933) / 1.372 =  2 moles of C

The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.

The molecular formula is Ag2C2O4

Explanation:

Step 1: Data given

A compound contains C, O and Ag

The mass of the compound is 1.372 grams

Mass of O = 0.288 grams

Mass of Ag = 0.974 grams

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass Ag =107.87 g/mol

The molar mass of the compound = 308.8 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles O = 0.288 grams / 16.0 grams

Moles O = 0.018 moles

Moles Ag = 0.974 grams / 107.87 g/mol

Moles Ag = 0.00903 moles

Moles C = 0.11 grams / 12.01 g/mol

Moles C = 0.00916 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

O: 0.018/ 0.00903 = 2

C: 0.0.00916 / 0.00903 = 1

Ag: 0.00903/0.00903 =

The empirical formula is AgCO2

The molar mass of this empirical formula is 151.88

We have to multiply the empirical formula by n

n = 308.8/151.88 = 2

2* (AgCO2) = Ag2C2O4

The molecular formula is Ag2C2O4

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g