Formation Equations - Online Tutor, Practice Problems & Exam Prep

Elements on the periodic table are found in different standard states at room temperature (approximately 25 degrees Celsius) and standard pressure (approximately one atmosphere), which include solids, liquids, and gases. Noble gases like helium and radon are monoatomic gases, while hydrogen ($H_2$), nitrogen ($N_2$), oxygen ($O_2$), fluorine ($F_2$), and chlorine (${\mathrm{Cl}}_2$) exist as diatomic gases. Bromine (${\mathrm{Br}}_2$) and mercury ($\mathrm{Hg}$) are the only liquid elements at room temperature, with bromine being diatomic and mercury monoatomic. Most other elements are solid and monoatomic in their natural state. Phosphorus ($P_4$), sulfur ($S_8$), selenium (${\mathrm{Se}}_8$), and tellurium (${\mathrm{Te}}_8$) exist as polyatomic molecules, while iodine ($I_2$) is diatomic. Elements with high atomic masses or those synthesized in laboratories may not have a defined phase under standard conditions. The molecular arrangement varies across phases: tightly packed in solids, more freedom of movement in liquids, and widely spread in gases. Understanding these states is essential for comprehending the physical properties and behavior of elements under different conditions.

Formation Equations

Realize that elements of the periodic table exist in different standard states or sometimes called natural states. When it comes to the natural world, we're going to say here at room temperature which is approximately 25�C, and standard pressure which is approximately 1 atmosphere. Elements exist as either solids, liquids, or gases.

So for a quick review on the periodic table, what we have here in red are the elements that exist as gases. Now some of them are monoatomic in terms of the noble gases from helium all the way down to radon. But other gases exist in a diatomic form where there are two of them combined together. This includes hydrogen, which is H2, nitrogen which is N2, oxygen which is O2, F2 for fluorine, and Cl2 for chlorine.

Then we have a few elements that can exist in the liquid phase. This includes bromine, which is also diatomic, and mercury, which is monoatomic. A vast majority of the other elements on the periodic table in their natural state or standard states exist in the solid form as monoatomic.

And then here we have in the bottom row these that aren't solids, liquids, or gases. These have been formed within laboratories or because of their high atomic masses are very unstable. So they don't come along with a phase at standard room temperature or pressure.

Now remember, in terms of these three phases in a solid, the molecules are very tightly packed together. In a liquid, they're more free to move around one another, and in gases, they're greatly spread apart. So just keep in mind when we're talking about the natural standard state of elements, these are how they exist in terms of phases and in terms of forms.

Now we also have phosphorus, sulfur, and selenium. These don't exist as monoatomic or diatomic atoms. Instead, they're going to exist as polyatomic molecules. Phosphorus exists as P4, sulfur is S8, and selenium and tellurium, which are below it, exist as Se8 and Te8. Iodine also is diatomic; it exists as I2. So keep in mind these different forms of the elements in their standard or natural states.

Formation Equations Example 1

Here it says which of the following elements is found in its standard state. If we take a look we have I₂. We know that I₂ iodine exists as a solid in its standard or natural state.

O₃ gas. Oxygen naturally exists as a diatomic molecule. It's O₂ gas, not O₃ gas. O₃ would be ozone. Next we have neon gas. Neon is a noble gas. Noble gases. Most of them are gases and they're monoatomic. This would represent its standard state.

And then here we have boron. Boron exists as a solid and it's monoatomic. It wouldn't exist as B₂ gas, so the only one in its standard state would be option C.

Formation Equations Example 2

In a formation equation, the standard state of elements combined together to form 1 mole of product. Here in this example it says write the formation equation for methane thiol which is chapter 3, SH. Following the steps, it says step one, write the compound as a product. So chapter 3 SH. In a formation equation, the product will always have a coefficient of one because remember, we're making one mole of it.

Step 2. Write the standard states of the elements that make up the element as the reactants. So it's made-up of carbon. Carbon is monoatomic, and the most common form of carbon is graphite plus hydrogen. Hydrogen is diatomic, and it's a gas plus sulfur. Sulfur is polyatomic. It's eight and it is a solid. Write the formation equation with the appropriate coefficients for the reactants.

A formation equation is a rare instance where a coefficient doesn't need a whole number. So if we were to balance this out, remember we make a list on both sides. So this is carbon, hydrogen and sulfur. Carbon, hydrogen and sulfur. We have one carbon, 2 hydrogens, 8 sulfurs, 1 carbon, 4 hydrogens, 1 sulfur. The coefficient here has to be a one. We cannot change that, so we're going to have to adjust the other coefficients.

Your carbons are tied for both 1:00 and 1:00, so that's fine. Here we have 4 hydrogens, but here we have two. So I'm going to put a 2 here. That means the two gets distributed to give us 4 hydrogens. Now then, we have 8 hydrogens here, but only one here. I'm not allowed to put a coefficient in front of the product, so this is one of those rare occasions where I'm going to have to do 18 because 18 * 8 will give me one. So now both sides of the equation have one solver involved and then we can just say there's a coefficient of one if you want in front of the carbon graphite.

So here this would be our balanced equation, 1 mole of carbon graphite +2 moles of hydrogen gas react with 18 sulfur 8 to produce one mole of methane thiol. So this would be our formation equation for methane thiol.

Which of the following represents the formation equation for barium nitrate, Ba(NO₃)₂?

• A. Ba²⁺ (aq) + 2 NO₃^– (aq) → Ba(NO₃)₂ (aq)
• B. Ba(NO₃)₂ (aq) → Ba (s) + N₂ (g) + 3 O₂ (g)
• C. Ba (s) + N₂ (g) + 3 O₂ (g) → Ba(NO₃)₂ (aq)
• D. Ba (s) + 2 N (g) + 6 O (g) → Ba(NO₃)₂ (aq)

Identify a substance that is not in its standard state.

• A. Cl₂
• B. Li
• C. H
• D. P₄
• E. Ar

Ibuprofen is used as an anti-inflammatory agent used to deal with pain and bring down fevers. If it has a molecular formula of C₁₃H­₁₈O₂, determine the balanced chemical equation that would give you directly the enthalpy of formation for ibuprofen. 

• A. 13 C (graphite) + H₂ (g) + O₂ (g) → C₁₃H­₁₈O₂ (s)
• B. 13 C (graphite) + 9 H₂ (g) + 2 O₂ (g) → C₁₃H­₁₈O₂ (s)
• C. 13 C (graphite) + 8 H₂ (g) + O₂ (g) → C₁₃H­₁₈O₂ (s)
• D. 13 C (graphite) + 9 H₂ (g) + O₂ (g) → C₁₃H­₁₈O₂ (s)