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3 years ago

Determine the empirical formula of a compound containing 48.38 grams of carbon, 8.12 grams of hydrogen, and 53.5 grams of oxygen

1 answer:

6 0

As with most stoichiometry problems, it is necessary to work in moles. The ratio of the moles of each element will provide the ratio of the atoms of each element.

Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).

Remeber that percentages are a ratio multiplied by 100. You must convert percentages back to their decimal value before working with them.

(.4838) (100 g) = 48.38 g C

(.0812 ) (100 g) = 8.12 g H

(.5350) (100 g) = 53.38 g O

Convert the mass of each element to moles of each element using the atomic masses.

(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C

(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H

(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O

Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.

Use the mole ratio to write the empirical formula.

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Explanation:

The nitrogen group is considered mixed because it is composed of elements of equally different classifications of metal, non-metal, and metalloids. The group contains the nonmetals nitrogen (N), and phosphorus (P), the metalloids arsenic (As) and antimony (Sb), and the metals bismuth (Bi) and Moscovium (Mc).

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3 years ago

Which combination of an element and an ion will react? View Available Hint(s) Which combination of an element and an ion will re

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Answer: The combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$ ......(1)

For the given options:

Option 1: $Sn(s)\text{ and }Mn^{2+}(aq.)$

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V$

Putting values in equation 1, we get:

$E^o_{cell}=-1.18-(-0.14)=-1.04V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 2: $Fe(s)\text{ and }Ca^{2+}(aq.)$

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$

Putting values in equation 1, we get:

$E^o_{cell}=-2.87-(-0.44)=-2.43V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 3: $Ni(s)\text{ and }Pt^{2+}(aq.)$

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V$

Reduction half reaction: $Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V$

Putting values in equation 1, we get:

$E^o_{cell}=1.2-(-0.25)=1.45V$

As, the standard potential is coming out to be positive, the given reaction will take place.

Option 4: $H_2(g)\text{ and }Na^{+}(aq.)$

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V$

Reduction half reaction: $Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V$

Putting values in equation 1, we get:

$E^o_{cell}=-0.27-(-0)=-0.27V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

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Answer:

Explanation:

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