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Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 15.6 g of carbon were burned in the presence of

NeX [460]

Answer: 5.72 g mass of carbon dioxide was produced

Explanation:

$C+O_2\rightarrow CO_2$

Moles of C = $\frac{\text{mass of carbon}}{\text{molar mass of carbon}}=\frac{15.6 g}{12 g/mol}=1.3 moles$

Mass of $O_2$ reacted = 59.1 g - 17.5 g = 41.6 g

Moles of $O_2\text{ reacted}=\frac{\text{mass of}O_2}{\text{molar mass of}O_2}=\frac{41.6 g}{32 g/mol}=1.3 moles$

According to reaction 1 mole $O_2$ produces 1 mole of $CO_2$ then 1.3 mole of $O_2$ will produce $\frac{1}{1}\times 1.3$ moles of $CO_2$

Mass of $CO_2$ :

=Moles of $CO_2$ × molar mass of $CO_2$ = 1.3 ×44 g/mol =

=57.2 grams

5.72 g mass of carbon dioxide was produced

7 0

2 years ago

A rocket travels 4500 meters in 150 seconds. What is its average speed?

Dmitrij [34]

Answer:

Your average speed was 3.77 Miles per hour

Explanation:

5 0

2 years ago

Arrange the elements in order of increasing atomic radlus. Use the periodic table

nlexa [21]

Largest to smallest: barium, calcium, selenium, oxygen

3 0

2 years ago

Read 2 more answers

10g of an oxide of copper contains 8.882g of copper. what is empirical formula of the copper oxide

-BARSIC- [3]

Answer:

Mass of copper oxide: 10g

Mass of copper: 8.882g

Mass of oxygen: 1.118g

8.882 1.118

Divide by $A_{r}$ 8.882÷63.5 1.118÷16

0.14 0.07

Divide by the lowest number 0.14/0.07 0.07/0.07

2 1

Check the ratio now, Cu has 2 atoms and O has 1 atom .

The formula is $Cu_{2} O$

4 0

3 years ago

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