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Aneli [31]
3 years ago
11

The foul odor of rancid butter is due largely to butyric acid, a compound containing carbon, hydrogen, and oxygen. combustion an

alysis of a 4.30-g sample of butyric acid produced 8.59 g co2 and 3.52 g h2o. part a find the empirical formula for butyric acid. express your answer as a chemical formula.
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
3 0

Mass of CO₂ = 8.59 g  

Molar mass of CO₂ = 44 g/mol

Moles of CO₂ = Mass of CO₂ / Molar mass of CO₂ = 8.59 g  / 44 g/mol

                        = 0.1952 moles

Now there is 1 mole of C in 1 mole of CO₂

Therefore, moles of carbon = moles of CO₂ = 0.1952 moles

Mass of carbon = moles of C x  the atomic weight of C

                           = 0.1952 moles x 12 g/mol

                            = 2.3427 g

 Mass of H₂O = 3.52g  

Molar mass of H₂O = 18g/mol

Moles of H₂O =  3.52g /18g/mol = 0.1956 moles

Now there is 2 moles of H in 1 mole of H₂O

Moles of H = 0.1956 moles X 2  = 0.3912 moles

Mass of H = 0.3912 moles X atomic mass of H  

                 = 0.3912 moles X  1 g/mol = 0.3912 g

Mass of butyric acid = 4.30g  

Mass of O = mass of butyric acid - (mass of C + mass of H)

                  = 4.30 g - 2.3427 g - 0.3912 g = 1.5661 g

Moles of O = mass of O / atomic mass of O = 1.5661 g / 16 g/mol

                   = 0.0978 moles  

Divide by smallest mole to get the molar ratio  

C = 0.1952 moles/0.0978 moles  = 2

H =  0.3912 moles/0.0978 moles  = 4

O = 0.0978 moles  / 0.0978 moles  = 1

Empirical formula of butric acid is C₂H₄O  

Molecular formula of butric acid is C₄H₈O₂


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3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To balance it, we can simply add another sodium atom on the left. Hence:

\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Given the initial value and the above ratios, this yields:

\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Cancel like units:

=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}

Multiply. Hence:

=0.3463...\text{ g H$_2$}

Since we should have two significant values:

=0.35\text{ g H$_2$}

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

Using the given value and the above ratios, we acquire:

\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

Cancel like units:

\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}

Multiply:

\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}

Since the resulting value should have three significant figures:

\displaystyle = 65.2 \text{ g Al$_2$O$_3$}

So, approximately 65.2 grams of aluminum oxide is produced.

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Explanation:

At STP, 1 mol of any gas will occupy 22.4 L.

With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:

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