Chemistry, 07.10.2019 20:00 hala201490

Be sure to answer all parts. dimercaprol (hsch2chshch2oh) was developed during world war i as an antidote to arsenic-based poison gas and is used today to treat heavy-metal poisoning. it binds the toxic element and carries it out of the body. (a) if each molecule binds one arsenic (as) atom, how many atoms of as could be removed by 696 mg of dimercaprol? × 10 atoms as (enter your answer in scientific notation.) (b) if one molecule binds one metal atom, calculate the mass % of each of the following metals in a metal-dimercaprol combination: mercury, thallium, chromium.

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Answer from: mariss8152

For A: The number of arsenic atoms are $3.4\times 10^{21}$

For B: The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Explanation:

For A:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of dimercaprol = 696 mg = 0.696 g (Conversion factor: 1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

$\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.0056 moles of dimercaprol will contain $0.0056\times 6.022\times 10^{23}=3.4\times 10^{21}$ number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, $3.4\times 10^{21}$ number of dimercaprol molecules will bind with = $1\times 3.4\times 10^{21}=3.4\times 10^{21}$ number of arsenic atoms

Hence, the number of arsenic atoms are $3.4\times 10^{21}$

For B:

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

$\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100$ ......(1)

For mercury:

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

$\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%$

For thallium:

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

$\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%$

For chromium:

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

$\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%$

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Answer from: Quest

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1

0.44 / 0.09 = 4.8

mgco3-5h2o"

Answer from: Quest

since electrons are negative the answer is neagtaive

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