A given sample of pure compound contains 9.81 g of Zn, 1.8 × 1023 atoms of chromium, and 0.60 mol of oxygen atoms. What is the simplest formula?

Question

A given sample of pure compound contains 9.81 g of Zn, 1.8 × 1023 atoms of chromium, and 0.60 mol of oxygen atoms. What is the simplest formula? (A) ZnCr2O7 (B) ZnCr2O4 (C) ZnCrO4 (D) ZnCrO6

Tardigrade Admin · Accepted Answer

Moles of Zn= (9.81/65.39)=0.15 Moles of Cr=(1.8×1023/6.023×1023)=0.3 Moles of O = 0.6 So, whole molar ratio Hence, formula = ZnCr2O4

Q. A given sample of pure compound contains $9.81 g$ of $Z n, 1.8 \times 1 0^{23}$ atoms of chromium, and $0.60 m o l$ of oxygen atoms. What is the simplest formula?

$Z n C r_{2} O_{7}$

$Z n C r_{2} O_{4}$

$Z n C r O_{4}$

$Z n C r O_{6}$

Moles of $Z n = \frac{9.81}{65.39} = 0.15$
Moles of $C r = \frac{1.8 \times 1 0 ^{23}}{6.023 \times 1 0 ^{23}} = 0.3$
Moles of $O = 0.6$
So, whole molar ratio
Hence, formula $= Z n C r_{2} O_{4}$

Q. A given sample of pure compound contains 9.81g of Zn,1.8×1023 atoms of chromium, and 0.60mol of oxygen atoms. What is the simplest formula?

ZnCr2​O7​

ZnCr2​O4​

ZnCrO4​

ZnCrO6​