Question

question 5
1 pts
what is the empirical formula for glucose?
c3h6o3
c6h12o6
ch2o
c2h4o2
question 6
1 pts
calculate the empirical formula for a compound comprised of 72.2% magnesium and 27.8% nitrogen by mass.
the following rhyme may be useful:
percent to mass
mass to mole
divide by small
multiply til whole
mg4n4
mg2n3
mg3n2
mgn

Explanation:

Question 5

The molecular formula of glucose is \(C_6H_{12}O_6\). The empirical formula is the simplest whole - number ratio of atoms in a compound. For \(C_6H_{12}O_6\), dividing each subscript by 6 (the greatest common divisor of 6, 12, and 6) gives \(CH_2O\).

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the mass of \(Mg = 72.2g\) and the mass of \(N=27.8g\)

Step2: Calculate the number of moles

The molar mass of \(Mg\) is \(M_{Mg}=24.31g/mol\), and the molar mass of \(N\) is \(M_{N} = 14.01g/mol\)
The number of moles of \(Mg,n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{72.2g}{24.31g/mol}\approx2.97mol\)
The number of moles of \(N,n_{N}=\frac{m_{N}}{M_{N}}=\frac{27.8g}{14.01g/mol}\approx1.98mol\)

Step3: Divide by the smallest number of moles

Dividing by \(n_{N}\) (since \(1.98<2.97\))
\(\frac{n_{Mg}}{n_{N}}=\frac{2.97}{1.98}\approx1.5\) and \(\frac{n_{N}}{n_{N}} = 1\)

Step4: Multiply to get whole - numbers

Multiply by 2 to get whole - numbers. So the ratio of \(Mg:N=(1.5\times2):(1\times 2)=3:2\)

Answer:

\(CH_2O\)

Question 6