Question

warm up – sept 17th orbital notation and % yield
1s 2s 2p 3s 3p
↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓
electron configuration of k: 1s² 2s²2p⁶ 3s²3p⁶ 4s¹

1. write the electron configuration and orbital notation of phosphorus.
2. in an experiment conducted by a student, 33 g h₂ reacts with 300.0 g o₂ to produce h₂o.

____h₂ + __ o₂ → ____h₂o
(a) calculate the theoretical yield or maximum amount in grams of h₂o produced in the reaction below.
? g h₂o = 33 g h₂ x
? g h₂o = 300.0 g o₂ x
(b) what is the percentage yield of this reaction?
% yield = actual yield / theoretical yield × 100%

Explanation:

Step1: Determine electron configuration of phosphorus

Phosphorus has atomic number 15. Electrons fill orbitals in order of increasing energy.
The electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$.

Step2: Write orbital notation of phosphorus

1s orbital: $\boxed{\uparrow\downarrow}$
2s orbital: $\boxed{\uparrow\downarrow}$
2p orbitals: $\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}$
3s orbital: $\boxed{\uparrow\downarrow}$
3p orbitals: $\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}$

Step3: Balance the chemical equation

The balanced equation for the reaction between hydrogen and oxygen to form water is $2H_{2}+O_{2}
ightarrow2H_{2}O$.

Step4: Calculate moles of reactants

The molar mass of $H_{2}$ is $M_{H_{2}} = 2.02\ g/mol$, so the number of moles of $H_{2}$, $n_{H_{2}}=\frac{33\ g}{2.02\ g/mol}=16.34\ mol$.
The molar mass of $O_{2}$ is $M_{O_{2}} = 32.00\ g/mol$, so the number of moles of $O_{2}$, $n_{O_{2}}=\frac{300.0\ g}{32.00\ g/mol}=9.375\ mol$.

Step5: Determine limiting reactant

From the balanced equation, the mole - ratio of $H_{2}$ to $O_{2}$ is 2:1.
For 9.375 mol of $O_{2}$, we would need $2\times9.375 = 18.75\ mol$ of $H_{2}$. But we have only 16.34 mol of $H_{2}$, so $H_{2}$ is the limiting reactant.

Step6: Calculate theoretical yield of $H_{2}O$

From the balanced equation, the mole - ratio of $H_{2}$ to $H_{2}O$ is 1:1.
The molar mass of $H_{2}O$ is $M_{H_{2}O}=18.02\ g/mol$.
The number of moles of $H_{2}O$ produced is equal to the number of moles of $H_{2}$ (since $H_{2}$ is limiting), so $n_{H_{2}O}=16.34\ mol$.
The theoretical yield of $H_{2}O$, $m_{H_{2}O}=n_{H_{2}O}\times M_{H_{2}O}=16.34\ mol\times18.02\ g/mol = 294.45\ g$.
Since we don't know the actual yield, we can't calculate the percentage yield. But if we assume the actual yield is given later, the formula for percentage yield is $\% \text{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$.

Answer:

1. Electron configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$

Orbital notation: 1s: $\boxed{\uparrow\downarrow}$, 2s: $\boxed{\uparrow\downarrow}$, 2p: $\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}$, 3s: $\boxed{\uparrow\downarrow}$, 3p: $\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}$

1. (a) Theoretical yield of $H_{2}O$ is 294.45 g

(b) Cannot be calculated without actual yield.