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3 years ago

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

Chemistry

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

Formulas

3.2 Determining Empirical and Molecular Formulas

Learning Objectives

By the end of this section, you will be able to:

Compute the percent composition of a compound

Determine the empirical formula of a compound

Determine the molecular formula of a compound

The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

%H=mass Hmass compound×100%

%C=mass Cmass compound×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

%H=2.5g H10.0g compound×100%=25%

%C=7.5g C10.0g compound×100%=75%

EXAMPLE 3.9

Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Check Your Learning

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

ANSWER:

12.1% C, 16.1% O, 71.8% Cl

Determining Percent Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the

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Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

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The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

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3 years ago

Sulfuryl chloride decomposes at high temperatures to produce sulfur dioxide and chlorine gases:

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Answer:

[SO2Cl2] = = 0.015 M [SO2] = = 0.0027 M [Cl2] = = 0.0027 M Q = = = 4.8 × 10−4

No. Q < Kc, so reaction will shift to the right.

Explanation:

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3 years ago

Atoms of arsenic (As) are often added to silicon (Si) in a process called doping to change the conductivity of the silicon. How

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Answer:

d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.

Explanation:

This is an example of a n-type semiconductor. The additional electron introduced to the 'grid' of silicon atoms causes an increase in the conductivity of the silicon. This additional electron is introduced as arsenic loses its extra electron.

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3 years ago

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Energy produced by nuclear fusion and fission

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3 years ago

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What is the molar mass of an unknown gas<br> with a density of 4.95 g/L at 1.00 atm and<br> 25.0 °C?

mestny [16]

Answer:

121 g/mol

Explanation:

To find the molar mass, you first need to calculate the number of moles. For this, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Because density is comparing the mass per 1 liter, I am assuming that the system has a volume of 1 L. Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 1.00 atm R = 0.0821 L*atm/mol*K

V = 1.00 L T = 25.0. °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(1.00 atm)(1.00L) = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = (24.478115)n

0.0409 = n

Now, we need to find the molar mass using the number of moles per liter (calculated) and the density.

0.0409 moles ? grams 4.95 grams
---------------------- x ------------------ = ------------------
1 L 1 mole 1 L

? g/mol = 121 g/mol

**note: I am not 100% confident on this answer

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2 years ago