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1) To get the empirical formula, we'll need to do the following:
First, we'll need to determine the mass of mercury (Hg) using the following equation:
[tex]mHg=mTotal-mBr=0,389 g - 0,111 g = 0.278 g Hg[/tex]
Now, we calculate the moles of each atom, using atomic masses:
[tex]molesHg=0,278gHg* \frac{1 mol Hg}{200,592gHg}= 0,001386 molHg \\ \\ molesBr=0,111gBr* \frac{1 mol Br}{79,904gBr}= 0,001389 molBr[/tex]
To finish, we express the formula using the moles, and dividing each amount by the lower one (0,001386 mol):
[tex]Hg _{ \frac{0,001386}{0,001386} } Br_{\frac{0,001389}{0,001386}} =Hg_1Br_1=HgBr[/tex]
2) To calculate the molecular formula, well need to do the following:
First, we calculate the molecular mass of the empirical formula:
[tex]MMHgBr= AM Hg + AM Br= 200,592 g/mol + 79,904 g/mol \\ \\ =280 gHgBr/mol[/tex]
Now, we divide the molecular mass of the compound between the molecular mass of the empirical formula to know how many empirical formulas are in the molecular formula:
[tex]\frac{561g/mol}{280g/mol}=2 [/tex]
To finish, we multiply the coefficients in the empirical formula by the number we just calculated to get the molecular formula:
[tex]Hg_{1*2} Br_{1*2}=Hg_2Br_2 [/tex]
Have a nice day!
1) To get the empirical formula, we'll need to do the following:
First, we'll need to determine the mass of mercury (Hg) using the following equation:
[tex]mHg=mTotal-mBr=0,389 g - 0,111 g = 0.278 g Hg[/tex]
Now, we calculate the moles of each atom, using atomic masses:
[tex]molesHg=0,278gHg* \frac{1 mol Hg}{200,592gHg}= 0,001386 molHg \\ \\ molesBr=0,111gBr* \frac{1 mol Br}{79,904gBr}= 0,001389 molBr[/tex]
To finish, we express the formula using the moles, and dividing each amount by the lower one (0,001386 mol):
[tex]Hg _{ \frac{0,001386}{0,001386} } Br_{\frac{0,001389}{0,001386}} =Hg_1Br_1=HgBr[/tex]
2) To calculate the molecular formula, well need to do the following:
First, we calculate the molecular mass of the empirical formula:
[tex]MMHgBr= AM Hg + AM Br= 200,592 g/mol + 79,904 g/mol \\ \\ =280 gHgBr/mol[/tex]
Now, we divide the molecular mass of the compound between the molecular mass of the empirical formula to know how many empirical formulas are in the molecular formula:
[tex]\frac{561g/mol}{280g/mol}=2 [/tex]
To finish, we multiply the coefficients in the empirical formula by the number we just calculated to get the molecular formula:
[tex]Hg_{1*2} Br_{1*2}=Hg_2Br_2 [/tex]
Have a nice day!
The empirical formula of the given compound is [tex]\boxed{{\text{HgBr}}}[/tex].
The molecular formula of the given compound is [tex]\boxed{{\text{H}}{{\text{g}}_2}{\text{B}}{{\text{r}}_2}}[/tex] .
Further explanation:
Empirical formula:
It is atom’s simplest positive integer ratio in compound. It may or may not be same as that of molecular formula. For example, the empirical formula of sulfur dioxide is [tex]{\text{SO}}[/tex] .
Molecular formula:
It is chemical formula that indicates total number and kinds of atoms in molecule. For example, molecular formula of sulfur dioxide is [tex]{\text{S}}{{\text{O}}_2}[/tex] .
Step 1: Mass of mercury (Hg) is to be calculated. This is done by using equation (1).
Since the given compound consists of only mercury and bromine. So the mass of mercury is calculated as follows:
[tex]{\text{Mass of mercury}}\left({{\text{Hg}}}\right)={\text{Mass of compound}} - {\text{Mass of bromine}}\left({{\text{Br}}}\right)[/tex]
…… (1)
The mass of the compound is 0.389 g.
The mass of bromine is 0.111 g.
Substitute the values in equation (1).
[tex]\begin{aligned}{\text{Mass of mercury}}&={\text{0}}{\text{.389 g}} - {\text{0}}{\text{.111 g}}\\&={\text{0}}{\text{.278 g}}\\\end{aligned}[/tex]
Step 2: The moles of mercury and bromine are to be calculated.
The formula to calculate the moles of mercury is as follows:
[tex]{\text{Moles of Hg}} = \frac{{{\text{Given mass of Hg}}}}{{{\text{Molar mass of Hg}}}}[/tex] …… (2)
The given mass of Hg is 0.278 g.
The molar mass of Hg is 200.59 g/mol.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{Moles of Hg}}&=\left( {{\text{0}}{\text{.278 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{200}}{\text{.59 g}}}}}\right)\\&=0.0013859\\&\approx{\text{0}}{\text{.001386 mol}}\\\end{gathered}[/tex]
The formula to calculate the moles of bromine is as follows:
[tex]{\text{Moles of Br}} = \frac{{{\text{Given mass of Br}}}}{{{\text{Molar mass of Br}}}}[/tex] …… (3)
The given mass of Br is 0.111 g.
The molar mass of Br is 79.90 g/mol.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Moles of Br}}&=\left({{\text{0}}{\text{.111 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{79}}{\text{.90 g}}}}}\right)\\&= 0.0013892\\&\approx{\text{0}}{\text{.001389 mol}}\\\end{gathered}[/tex]
Step 4: The moles of mercury and bromine are to be written with their corresponding subscripts.
So the preliminary formula becomes,
[tex]{\text{Preliminary formula of the compound}}={\text{H}}{{\text{g}}_{0.001386}}{\text{B}}{{\text{r}}_{0.001389}}[/tex]
Step: Each of the subscripts is divided by the smallest subscript to get the empirical formula.
In this case, the smallest one is 0.001386. So the empirical formula of the compound is written as follows:
[tex]\begin{aligned}{\text{Empirical formula of the compound}}&={\text{H}}{{\text{g}}_{\frac{{0.001386}}{{0.001386}}}}{\text{B}}{{\text{r}}_{\frac{{0.001389}}{{0.001386}}}}\\&={\text{H}}{{\text{g}}_1}{\text{B}}{{\text{r}}_{1.002}}\\&\approx {\text{HgBr}}\\\end{aligned}[/tex]
Therefore the empirical formula of the compound is [tex]{\mathbf{HgBr}}[/tex] .
Step 6: The empirical formula mass of the compound is to be calculated. This is done by using equation (4).
[tex]{\text{Empirical formula mass of HgBr}}=\left( 1 \right)\left({{\text{Atomic mass of Hg}}}\right) + \left(1\right)\left({{\text{Atomic mass of Br}}}\right)[/tex]
…… (4)
Substitute 200.59 g/mol for the atomic mass of Hg and 79.90 g/mol for the atomic mass of Br in equation (4).
[tex]\begin{aligned}{\text{Empirical formula mass of HgBr}}&=\left( 1 \right)\left({{\text{200}}{\text{.59 g/mol}}}\right) + \left( 1 \right)\left( {{\text{79}}{\text{.90 g/mol}}}\right)\\ &= 280.49\;{\text{g/mol}}\\\end{gathered}[/tex]
Step 7: The molar mass of the compound is divided by its empirical formula mass to get a whole number. The formula for this is as follows:
[tex]{\text{Whole - number multiple}} = \frac{{{\text{Molar mass of compound}}}}{{{\text{Empirical formula mass}}}}[/tex] …… (5)
Substitute 560 g/mol for the molar mass of the compound and 280.49 g/mol for the empirical formula mass of the compound in equation (5).
[tex]\begin{aligned}{\text{Whole - number multiple}}& = \frac{{{\text{560 g/mol}}}}{{{\text{280}}{\text{.49 g/mol}}}}\\&= 1.99\\&\approx2\\\end{aligned}[/tex]
Step 8: The empirical formula is multiplied by the whole number multiple to get the molecular formula. So the molecular formula of the compound is [tex]{\mathbf{H}}{{\mathbf{g}}_{\mathbf{2}}}{\mathbf{B}}{{\mathbf{r}}_{\mathbf{2}}}[/tex] .
Learn more:
1. Calculate the moles of ions in the solution: https://brainly.com/question/5950133
2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Stoichiometry of formulas and equations
Keywords: empirical formula, Hg, Br, HgBr, Hg2Br2, subscript, moles of Br, moles of Hg, mass of Hg, mass of Br, molecular formula, 2, preliminary formula.
