Chemistry, 09.07.2019 21:00 balwinderdev

A0.999-g sample of a metal chloride, mcl2, is dissolved in water and treated with excess aqueous silver nitrate. the silver chloride that formed weighed 1.286 g. calculate the atomic mass of m.

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Answer from: 19colemankl

Mass of silver chloride formed = 1.286 g

Molar mass of silver chloride = 143.32 g/mol AgCl

Converting mass to moles of AgCl using the molar mass:

$1.286 g AgCl *\frac{1mol}{143.32g}= 0.00897 mol AgCl$

The balanced chemical equation between metal chloride and silver nitrate can be represented as,

$MCl_{2}(aq)+2AgNO_{3}(aq)--2AgCl(s)+M(NO_{3})_{2}(aq)$

Calculating the moles of MCl2 from moles of AgCl that form:

$0.00897 mol AgCl*\frac{1molMCl_{2} }{2molAgCl} =0.00449mol MCl_{2}$

The given mass of $MCl_{2}$ =0.999 g

Molar mass of $MCl_{2}$ = $\frac{0.999g}{0.00449mol} =222.5g/mol$

Molar mass of M= Molar mass of $MCl_{2}$ - 2 (Molar mass of Cl)

=151.6g/mol

Answer from: khalilh1206

Atomic mass of M is $\boxed{151.6\text{ g}}$ .

Further Explanation:

Given reaction occurs as follows:

$\text{MCl}_2(aq)+2\text{AgNO}_3(aq)\rightarrow\text{2AgCl}(s)+\text{M}\left(\text{NO}_3}\right)_2(aq)$

The formula to calculate moles of AgCl is as follows:

$\text{Moles of AgCl}=\dfrac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}}$ ...... (1)

Substitute $1.286\text{ g}$ for mass of AgCl and $143.32\text{ g/mol}$ for molar mass of AgCl in equation (1).

$\begin{aligned}\text{Moles of AgCl}=&\dfrac{\text{1.286 g}}{\text{143.32 g/mol}}\\=&0.00897\text{ mol}\end{aligned}$

According to the balanced chemical reaction, one mole of $\text{MCl}_2$ reacts with two moles of $\text{AgNO}_3$ to form two moles of AgCl and one mole of $\text{M}\left(\text{NO}_3\right)_2$ , So stoichiometric ratio between $\text{MCl}_2$ and AgCl is 1:2.

Since two moles of AgCl are produced by one mole of $\text{MCl}_2$ , number of moles of $\text{MCl}_2$ that can produce 0.00897 moles of AgCl can be calculated as follows:

$\begin{aligned}\text{Moles of MCl}_2=&\left(\dfrac{\text{1 mol of MCl}_2}{\text{2 mol of AgCl}}\right)\left(0.00897\text{mol AgCl}\right)\\=&0.00449\text{ mol}\end{aligned}$

The formula to calculate number of moles of $\text{MCl}_2$ is as follows:

$\text{Moles of MCl}_2=\dfrac{\text{Mass of MCl}_2}{\text{Molar mass of MCl}_2}$ ...... (2)

Rearrange equation (2) to calculate molar mass of $\text{MCl}_2$ . ${\text{Molar mass of MCl}_2}=\dfrac{\text{Mass of MCl}_2}{\text{Moles of MCl}_2}$ ...... (3)

Substitute 0.999 g for mass of $\text{MCl}_2$ and 0.00449 mol for moles of $\text{MCl}_2$ in equation (3).

$\begin{aligned}{\text{Molar mass of MCl}_2}=&\dfrac{\text{0.999 g}}{\text{0.00449 mol}}\\=&222.5\text{ g/mol}\end{aligned}$

The formula to calculate molar mass of $\text{MCl}_2$ is as follows:

$\text{Molar mass of MCl}_2=\left[1\left(\text{Atomic mass of M}\right)+\\\\2\left(\text{Atomic mass of Cl}\right)\right]}$ ....... (4)

Rearrange equation (4) to calculate atomic mass of M.

$\text{Atomic mass of M}=\left[\text{Molar mass of MCl}_2-2\left(\text{Atomic mass of Cl}\right)\right]$ ...... (5)

Substitute 222.5 g/mol for molar mass of $\text{MCl}_2$ and 35.453 g for atomic mass of Cl in equation (5).

$\begin{aligned}\text{Atomic mass of M}=&222.5\text{ g}-2\left(35.453\text{ g}\right)\\=&151.6\text{ g}\end{aligned}$

Therefore atomic mass of M is 151.6 g.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. Calculate the moles of ions in the solution:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: AgCl, MCl2, 151.6 g, atomic mass, molar mass, 222.5 g/mol, 35.453 g, 1.286 g, 0.999 g, stoichiometric ratio, 1:2.

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.