7.1.2: Practice Mole Calculations
- Page ID
- 236040
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Molar Mass, Gram to Mole, Mole to Gram
Exercise \(\PageIndex{1}\)
How many moles of Ca3(PO4)2 are in 57.82 grams?
- Molar Mass
-
Ca3(PO4)2 = \(\dfrac{310.18 g}{mol}\)
- Answer
-
0.1864 moles
How many grams of Zn(OH)2 are in 3.67 moles?
- Molar Mass
-
Zn(OH)2 = \(\dfrac{99.41 g}{mol}\)
- Answer
-
365 grams
How many moles of Mg(NO3)2 are in 487.6 grams?
- Molar Mass
-
Mg(NO3)2 = \(\dfrac{148.33 g}{mol}\)
- Answer
-
3.287 moles
Using Avogadro's Number
Exercise \(\PageIndex{1}\)
How many molecules of H2O are in 0.0643 grams of H2O?
- Answer
-
0.0643 grams of H2O is equal to 0.00357 moles, which is 2.15 x 1021 molecules
How many moles of Au is equal to 4.77 x 1027 Au atoms?
- Answer
-
7920 moles
Using Formula Subscripts
Exercise \(\PageIndex{1}\)
How many moles of NH3 would contain 8.18 moles of H atoms?
- Answer
-
2.73 moles NH3
How many Br atoms are in 0.453 moles of CBr4?
- Answer
-
1.09 x 1024 Br atoms
Mixed
Exercise \(\PageIndex{1}\)
How many moles of Ca(OH)2 are in 5.62 g of Ca(OH)2?
- Answer
-
0.0758 mol Ca(OH)2
How many moles of O are in this amount?
- Answer
-
0.152 mol O
How many individual atoms of O are in this amount?
- Answer
-
9.13 x 1022 O atoms
Exercise \(\PageIndex{1}\)
How many moles of C4H8O2 can be made with 18.5 mol C?
- Answer
-
4.63 moles of C4H8O2
Exercise \(\PageIndex{1}\)
How many grams of C4H8O2 can be made with 2.86 x 1021 C atoms? (Hint: how many moles of C is this?)
- Hint
-
0.00475 mol C
- Answer
-
0.00119 mol C4H8O2 which is 0.105 g C4H8O2