CHEMISTRY COLLEGE

What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

Answers

Answer 1
Answer:

Asnwer : Empirical formula of a compound is : C_(3)H_(4)O

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

(100g)* ((64.3percent))/((100percent)) = 64.3g

Mass of H = 7.2 g

(100g)* ((7.2percent))/((100percent)) = 7.2g

Mass of O = 28.5 g

(100g)* ((28.5percent))/((100percent)) = 28.5g

Step 2 : Convert the grams of each compound to moles.

Moles = (Grams)/(Molar mass)

Molar mass of C = 12.0g/mol  

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

Moles of C = (64.3g)/(12.0(g)/(mol))

Moles of C = 5.36 mol

Moles of H = (7.2g)/(1.0(g)/(mol))

Moles of H = 7.2 mol

Moles of O = (28.5g)/(16.0(g)/(mol))

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

Mole of C = (5.36mol)/(1.78mol) = 3.0

Mole of H = (7.2mol)/(1.78mol) = 4.0

Mole of O = (1.78mol)/(1.78mol) = 1.0

C : H : O = 3 : 4 : 1

So empirical formula of the compound is C_(3)H_(4)O_(1) or C_(3)H_(4)O

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Explanation:

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Answers

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Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answers

Answer:

\boxed{\text{23.4 mmHg}}

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

K_{\text{p}} = p_{\text{H2O}}

\text{The relationship between $\Delta G^(\circ)$ and $K_{\text{ p}}$ is}\n\Delta G^(\circ) = -RT \ln K_{\text{p}}

Data:  

T = 25 °C

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Calculations:

T = (25 + 273.15) K = 298.15 K

\begin{array}{rcl}8600 & = & -8.314 * 298.15 \ln K \n8600 & = & -2478.8 \ln K\n-3.47 & = & \ln K\nK&=&e^(-3.47)\n& = & 0.0311\end{array}

Standard pressure is 1 bar.

p_{\text{H2O}} = \text{0.0311 bar} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\n\n\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}

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Answers

Answer:

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Explanation:

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Answers

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Answers

Answer:

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