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2 years ago

Determine the mass in grams of each element.

Chemistry

1 answer:

jeka57 [31]2 years ago

7 0

1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g

2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g

3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g

4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g

<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>

Mole of Sb = 1.33×10²² mole

Molar mass of Sb = 122 g/mol

Mass of Sb =?

Mass = mole × molar mass

Mass of Sb = 1.33×10²² × 122

Mass of Sb = 1.62×10²⁴ g

<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>

Mole of Pt = 4.75×10¹⁴ mole

Molar mass of Pt = 122 g/mol

Mass of Pt =?

Mass = mole × molar mass

Mass of Pt = 4.75×10¹⁴ × 195

Mass of Pt = 9.26×10¹⁶ g

<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>

Mole of Ag = 1.22×10²³ mole

Molar mass of Ag = 108 g/mol

Mass of Ag =?

Mass = mole × molar mass

Mass of Ag = 1.22×10²³ × 108

Mass of Ag = 1.32×10²⁵ g

<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>

Mole of Cr = 9.85×10²⁴ mole

Molar mass of Cr = 52 g/mol

Mass of Cr =?

Mass = mole × molar mass

Mass of Cr = 9.85×10²⁴ × 52

Mass of Cr = 5.12×10²⁶ g

Learn more about mole:

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2 years ago

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

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Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

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Answer:

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<em>The overall reaction will be:</em>

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