Gaseous chlorine, Cl2, was combined with 0.532g of Titanium, and 2.108 g of TixCly were collected. What is the empirical formula of titanium chloride?

Answers

Answer 1

Answer:

TiCl₄

Explanation:

From the law of conservation of mass, the mass of the reactants must equal the mass of the products. In other words:

[tex]\displaystyle \text{m Cl$_2$} + \text{ m Ti} = \text{ m Ti$_x$Cl$_y$}[/tex]

Find the mass of chlorine gas reacted:

[tex]\displaystyle \begin{aligned} \text{ m Cl$_2$} + (0.532 \text{ g}) & = (2.108 \text{ g)} \\ \\ \text{m Cl$_2$} & = 1.576 \text{ g} \end{aligned}[/tex]

Find the number of moles of each reactant using their respective molecular weights:

[tex]\displaystyle 1.576 \text{ g Cl$_2$} \cdot \frac{1 \text{ mol Cl$_2$}}{70.90\text{ g Cl$_2$}} \cdot \frac{2\text{ mol Cl}}{1\text{ mol Cl$_2$}} = 0.04446\text{ mol Cl}[/tex]

And:

[tex]\displaystyle 0.532\text{ Ti} \cdot \frac{1\text{ mol Ti}}{47.87\text{ g Ti}} = 0.0111\text{ mol Ti}[/tex]

To find the empirical formula, all values by the smallest value:

[tex]\displaystyle \begin{aligned} \frac{0.0111\text{ mol Ti}}{0.0111} & = \frac{0.04446\text{ mol Cl}}{0.0111} \\ \\ 1\text{ mol Ti} & \approx 4\text{ mol Cl}\end{aligned}[/tex]

Hence, there are four moles of chlorine for every one mole of titanium.

In conclusion, our empirical formula is TiCl₄.


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Answers

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Answers

[tex]\huge \bf༆ Answer ༄[/tex]

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Answers

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A sample of tin (Cp = 0. 227 J/g•°C) is placed in a freezer. Its temperature decreases from 15. 0°C to â’10. 0°C as it releases 543 J of energy. What is the mass of the sample? Round your answer to three significant figures. G.

Answers

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Answers

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Answers

Answer:

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Answers

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Answers

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Answers

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