Question

1. find the mole weight of:

a. lioh
b. mgbr₂
c. ca(no₃)₂
d. al₂(co₃)₃

Explanation:

Step1: <Find atomic weights>

Atomic weights: \(Li = 6.94\ g/mol\), \(O = 16.00\ g/mol\), \(H = 1.01\ g/mol\), \(Mg = 24.31\ g/mol\), \(Br = 79.90\ g/mol\), \(Ca = 40.08\ g/mol\), \(N = 14.01\ g/mol\), \(Al = 26.98\ g/mol\), \(C = 12.01\ g/mol\)

Step2: <Calculate for \(LiOH\)>

\(MW_{LiOH}=6.94 + 16.00+1.01\)
\(MW_{LiOH}=23.95\ g/mol\)

Step3: <Calculate for \(MgBr_{2}\)>

\(MW_{MgBr_{2}}=24.31+(2\times79.90)\)
\(MW_{MgBr_{2}}=24.31 + 159.80\)
\(MW_{MgBr_{2}}=184.11\ g/mol\)

Step4: <Calculate for \(Ca(NO_{3})_{2}\)>

\(MW_{Ca(NO_{3})_{2}}=40.08+2\times(14.01 + 3\times16.00)\)
\(MW_{Ca(NO_{3})_{2}}=40.08+2\times(14.01+48.00)\)
\(MW_{Ca(NO_{3})_{2}}=40.08+2\times62.01\)
\(MW_{Ca(NO_{3})_{2}}=40.08 + 124.02\)
\(MW_{Ca(NO_{3})_{2}}=164.10\ g/mol\)

Step5: <Calculate for \(Al_{2}(CO_{3})_{3}\)>

\(MW_{Al_{2}(CO_{3})_{3}}=2\times26.98+3\times(12.01+3\times16.00)\)
\(MW_{Al_{2}(CO_{3})_{3}}=53.96+3\times(12.01 + 48.00)\)
\(MW_{Al_{2}(CO_{3})_{3}}=53.96+3\times60.01\)
\(MW_{Al_{2}(CO_{3})_{3}}=53.96+180.03\)
\(MW_{Al_{2}(CO_{3})_{3}}=233.99\ g/mol\)

Answer:

a. \(23.95\ g/mol\)
b. \(184.11\ g/mol\)
c. \(164.10\ g/mol\)
d. \(233.99\ g/mol\)