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2 years ago

Can someone please explain how to find an emperical formula given mass not percentage?

Chemistry

1 answer:

Firdavs [7]2 years ago

8 0

Answer:

1. Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages). ...

2. Convert the mass of each element to moles. ...

3. Find the ratio of the moles of each element. ..

4. Use the mole ratio to write the empirical fomula

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Explanation:

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3 years ago

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Answer:

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3 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

Sodium metal reacts with chlorine gas to produce sodium chloride. What mass (grams) of sodium metal would be needed to fully rea

gulaghasi [49]

Answer:

105.8 g of Na would be required

Explanation:

Let's think the reaction:

2Na(s) + Cl₂(g) → 2NaCl (s)

1 mol of chlorine reacts with 2 moles of sodium

Then, 2.3 moles of Cl₂ would react with (2.3 .2) / 1 = 4.6 moles

Let's determine the mass of them.

4.6 mol . 23 g/mol = 105.8 g

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3 years ago

Compare la configuración realizada con la que aparece en la tabla periódica y confirma si esta en

mart [117]

Answer:

Explanation:

4 0

3 years ago

Can someone please explain how to find an emperical formula given mass not percentage?​

Can someone please explain how to find an emperical formula given mass not percentage?