Question

can you do it backwards?
how many formula units will be in 89.3 grams of manganese (iv) carbonate? (use e instead of x10)
enter the number only this time. proper significant digits. no units. no formulas.

Explanation:

Step1: Calculate molar mass of Mn(CO₃)₂

Molar mass of Mn = 54.94 g/mol, C = 12.01 g/mol, O = 16.00 g/mol.
Molar mass of Mn(CO₃)₂ = 54.94+(2×(12.01 + 3×16.00)) = 54.94+(2×(12.01+48.00)) = 54.94+(2×60.01) = 54.94 + 120.02 = 174.96 g/mol

Step2: Calculate moles of Mn(CO₃)₂

Moles \(n=\frac{m}{M}\), where \(m = 89.3\) g and \(M=174.96\) g/mol.
\(n=\frac{89.3}{174.96}\approx0.510\) mol

Step3: Calculate formula units

Number of formula units \(N=n\times N_A\), where \(N_A = 6.022E23\) formula units/mol.
\(N=0.510\times6.022E23 = 3.07122E23\approx3.07E23\)

Answer:

3.07E23