Q.16 - An Introduction to General, Organic, and Biological Chemistry

Question

a. If black spheres are carbon atoms, white spheres are hydrogen atoms, and red spheres are oxygen atoms, what is the molecular formula of butyric acid?

b. What is the molar mass of butyric acid?

c. How many grams of butyric acid contain $3.28\times {10}^{23}$atoms of oxygen?

d. How many grams of carbon are in 5.28g of butyric acid?

e. Butyric acid has a density of 0.959 g/ml. How many moles of butyric acid are contained in 1.56 ml of butyric acid?

f. Identify the bonds C-C, C-H, and C-O in a molecule of butyric acid as polar covalent or nonpolar covalent.

Step-by-Step Solution

Answer

Part a. ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$

Part b. $88.104 \mathrm{g}/\mathrm{mole}$

Part c. $48 g$

Part d. $2.88 g$

Part e. $0.017$mole

Part f. C-C, C-H are nonopolar covalent and C-O is polar covalent

Step 1: Given Information (Part a)

A chemical bond keeps particles intact in atoms. Bonds emerge from the electrostatic powers between emphatically charged nuclear cores and adversely charged electrons (the places of which in not set in stone by quantum mechanics).

Step 2: Explanation Part (a)

Given,

The black spheres are carbon atoms, white spheres are hydrogen atoms, and red spheres are oxygen atoms.

Hence the molecular formula of butyric acid is ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$.

Step 3: Explanation Part (b)

The molar mass of ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$is,

C= $12.01 g$

H= $1.008 g$

O= $16 g$

$$\begin{gathered} =(4\times 12.01 \mathrm{g} \mathrm{C})+(8\times 1.008 \mathrm{g} \mathrm{H})+(2\times 16 \mathrm{g} \mathrm{O}) \\ =88.104 g \end{gathered}$$

Step 4: Explanation Part (c)

We know the molar mass of $C_4{H}_8{O}_2$ = $6.02\times {10}^{23}$atoms of $C_4{H}_8{O}_2$

The number of grams of butyric acid that contain $3.28\times {10}^{23}$atoms is,

$$\begin{gathered} 3.28\times {10}^{23}\text{ atom }\times \frac{1\text{ mole }}{6.02\times {10}^{23}\text{ atom }}\mathrm{x}\frac{88.104\mathrm{g}}{1\text{ mole }} \\ =48 \mathrm{g} \end{gathered}$$

Step 5: Explanation Part (d)

Calculating the number of moles butyric acid,

number of moles of butyric acid = $\frac{5.28 \mathrm{g}}{88.104 \mathrm{g}/\mathrm{mole}}=0.06 moles$

number of moles of carbon atom C = $0.06\text{ mole }{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2\times \frac{4\text{ mole }\mathrm{C}}{1\text{ mole }{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2}=0.24\text{ moles }$

The amount of carbon is,

$0.24\text{ mole }\times \frac{12.01 \mathrm{g} \mathrm{C}}{1 \mathrm{mole} \mathrm{C}}=2.88 \mathrm{g}$

Step 6: Explanation Part (e)

We know,

The density of butyric acid = $0.959 \mathrm{g}/\mathrm{ml}$

Volume= $1.56 ml$

Mass = Density$\times$Volume

Mass = $0.959\frac{\mathrm{g}}{\mathrm{mL}}\times 1.56 \mathrm{mL}=1.5 \mathrm{g}$

Number of moles = $\frac{amount in grams}{molar mass}$

Number of moles $=\frac{1.50 \mathrm{g}}{88.104 \mathrm{g}/\mathrm{mole}}=0.017 moles$

Step 7: Explanation Part (f)

The connection between C-C, and C-H are instances of nonpolar bonds since here the electrons of the covalent bond share similarities between the two atoms. The connection between C-O is an illustration of a polar bond since here the electrons of the covalent bond are shared inconsistent between two atoms. O is more electronegative than C, so O will draw in the common sets of electrons in the covalent bond towards itself, hence making the C-O tie polar.