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Vaselesa [24]
2 years ago
11

Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?

Chemistry
1 answer:
tester [92]2 years ago
5 0

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the <em>empirical formula </em>

EF = K₂O

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7 0
3 years ago
Read 2 more answers
If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an
kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

\displaystyle \frac{P}{RT} = \frac{n}{V}

The left-hand side is simply some constant. Hence, we can write that:

\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}

Substitute in known values:

\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})}  = \frac{(13.15\text{ mol })}{V_2}

Solving for <em>V</em>₂ yields:

\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0
1 year ago
so they are asking me to find the empirical formula for a compound that is 7.70% carbon and 92.3% chlorine. Can you show me step
Nikolay [14]

Answer:

Empirical formula is CCl₄

Explanation:

Given data:

Percentage of carbon = 7.70%

Percentage of chlorine = 92.3%

Empirical formula = ?

Solution:

Number of gram atoms of Cl = 92.3 / 35.5 = 2.6

Number of gram atoms of C = 7.70 / 12 = 0.64

Atomic ratio:

            C                      :              Cl            

           0.64/0.64        :             2.6/0.64

            1                      :                4        

C : Cl  = 1 : 4

Empirical formula is CCl₄.

5 0
3 years ago
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