A compound contains 16.7 g of iridium and 10.3 g of selenium, what is its empirical formula?

THE EMPIRICAL FORMULA OF THE COMPOUND IS Ir Se2

Explanation:

Iridium = 16.7 g

Selenium = 10.3 g

Total mass = 16.7 + 10.3 = 27 g

To calculate the empirical formula, we first calculate the percentage com[position of the individual elements.

Percentage composition of iridium = 16.7 / 27 * 100 = 61.85 %

Percentage composition of selenium = 10.3 / 27 * 100 = 38.15 %

Next is to divide the percentage composition of each element by their respective molecular masses

Molecular mass of iridium = 192 g/mol

Molecular mass of selenium = 79g/mol

Iridium = 61.85 / 192 =  0.3221

Selenium = 38.15 / 79 = 0.4829

Next is to divide the values by the smaller of the two values

Iridium = 0.3221 / 0.3221 = 1

Selenium = 0.4829 / 0.3221 = 1.5

Next is to round up the values to a whole number

Iridium = 1

Selenium = 2

The empirical formula therefore is Ir Se2

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6