Acompound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. part a what are the subscripts in the empirical formula of this compound? enter the subscripts for c, h, and o, respectively, separated by commas (e. g., 5,6,7).

Answer : The subscripts for C, H and O in the empirical formula of this compound is 1, 2 and 1

Solution : Given,

Mass of C = 40.0 g

Mass of H = 6.70 g

Mass of O = 53.3 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =

Moles of H =

Moles of O =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O =

The ratio of C : H : O = 1 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =

Therefore, the subscripts for C, H and O in the empirical formula of this compound is 1, 2 and 1

i don,t know so just answer mine

b) nuclear ive learned that in a pedtrodic table