8: Empirical Formulas And Hydrates
- Page ID
- 506186
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The purpose of this experiment is to:
- Determine the empirical formula of a hydrate.
- Calculate the percentage of water in a hydrate.
INTRODUCTION
An empirical formula is the smallest whole-number ratio of elements in a compound. It can be determined experimentally by examining a substance's molar ratio of individual components. Consider the following example:
A compound was shown to consist of 0.750 g of carbon and 2.000 g of oxygen. What is the empirical formula?
Solution
To calculate the empirical formula, we would first need to convert grams of each element to moles, as shown below:
Since we know the ratio of moles, we can automatically determine the ratio of atoms, as 1 mole equals 6.02 × 1023 atoms. Therefore, the ratio of moles of the carbon oxide compound is:
Since there are twice as many moles of oxygen as carbon, the empirical formula is CO2.
In this experiment, you will determine the empirical formula of a hydrate, a compound containing water molecules. First, you will heat the hydrate to drive off the water molecules and then determine the mass of water present in the hydrate by subtracting the mass of the anhydrous compound from the mass of the original hydrate. The empirical formula will be found by comparing the moles of anhydrous compound to the moles of water, as shown in the example below:
2.500 g of a copper(II) chloride hydrate was heated, resulting in 1.972 g of anhydrous compound. Using this information, determine the empirical formula of the copper chloride hydrate.
Solution
First, we can find the mass of water driven off:
2.500 g hydrate – 1.972 g anhydrous compound 0.528 g water
Next, we can find the molar ratio of the anhydrous compound (AC) to water. Since the anhydrous compound has the formula CuCl2, we can convert grams to moles, as follows:
Similarly, we can find the moles of water:
Comparing the molar ratios gives:
The empirical formula would therefore be written as CuCl2· 2H2O, indicating that there are two water molecules for every molecule of the anhydrous compound. Since the molar mass of the hydrate is 170.45 g/mol (adding up the molar masses from Cu, 2 Cl, and 2 H2O), the percentage of water can be calculated as:
