Question

Attempt 12 Question 1 of 3 > An unknown compound contains only C, H, and 0. Combustion of 6.60 g of this compound produced 15.0 g Co, and 6.14 g H,O. What is the empirical formula of the unknown compound? Insert subscripts as needed. empirical formula: CHO

Answer 1

Ans - let us assume that we have one more of all reagents and products. CxHy be the emperical formula of the unknown compound. The combustion reaction can be written "Ca HyO₂ + O2 CO2 + 20 I mole of a contains (12.01 + 2x15.99) g of CO2 - 43.99 g of loa ie 43.99 g contains i mole of co2 :.15g of cop contains I x 15 mole of co n = 0.34 moes of co 43.99 One compound of co, is made up of one carbon atom and two oxygen atoms. So, if we have 0.34 moles of coz then we have 0.34 moes of carbon in our unknown compound since we assume that one more of product. i man of carbon in untenown compound = 0.34 x 12.01g = 4.08g of c I mole of 120 contains (1.008 x 2 + 15.99) g of the - 18 g of 140 ne 18g of the contains I mile of 40 .. :.6.149 will contain 1/3 x 6.14 moles of the = 0.34 moles of 150

One compound of ho is made up of one oxygen atom and two hydrogen atoms. So if we have 0.34 mors of ho then we have (2x0.34) moles in 0.68 mous of hydrogen in the unknown compound. .. man of hydrogen in the unknown compound = 0.68 x 1.008 g = 0.685 g of Hydrogen. Adding the man of c and H will give us the man make up by c and H in the compound .. mas made up by (and H=0.685 +4.089 = 4.7659 So this means that the remaining mass is oxygen mass of oxygen = 6.60 -4.765 g = 1.8359. : number of molus of origin = 1.835 mous 15.99 = 0.11 miles Now, dividing by the smallest male value, we get 0134 - 3.09 3. 0.11 034 0.11 3,0923 0.11 / 0.11

The mlan salis of C#:0 in :3:1 Thus the empirical formula of the compound is Catho