-3
$\begingroup$

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass? If the law is true, shouldn't all chemical reactions balance the atoms in every single equation? It's like you would add 2 balls together and come up with 1 + 1 = 3 and then say that we "need to balance the equation" so it will make sense. Hopefully my question is clear.

$\endgroup$

closed as unclear what you're asking by Mithoron, A.K., airhuff, user55119, Mathew Mahindaratne Apr 28 at 5:12

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 8
    $\begingroup$ I'm sorry, but I have read your question five times, and probably will read it more often, but I don't have a clue what you are asking. $\endgroup$ – Martin - マーチン Apr 27 at 11:03
  • 2
    $\begingroup$ You've most likely misunderstood the law or its implications on chemical equations. I would suggest reading this introductory article. A chemical reaction does not produce an unbalanced equation, but we need to balance equations we create to match the stochiometry of the real reaction. $\endgroup$ – William R. Ebenezer Apr 27 at 11:38
  • $\begingroup$ You are indeed balancing equation according to the conservation of mass and the chemistry that is going on. And furthermore there are not unbalanced equation by definition (though balancing equation is surely commonly said). $\endgroup$ – Alchimista Apr 27 at 12:18
  • $\begingroup$ Could you provide an example reaction? $\endgroup$ – A.K. Apr 28 at 1:30
3
$\begingroup$

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass?

You find unbalanced equations as exercises and as a step in writing a balanced equation when there is insufficient information that still needs to be discovered.

Balancing equations as an exercise

The observable chemistry underlying a chemical equation is always balanced. Presenting an unbalanced equation and asking students to balance it is an exercise. Similar exercises are given in many other disciplines:

Balance the following equations:

$$\tag{Maths} (a + b)^2 = a^2 + ab + b^2$$

$$\tag{Physics}E = m c$$

$$\tag{Genetics}\text{Fruitfly = Heads and Legs and Wings and Eyes}$$

$$\tag{Accounting}\text{Dollar = Nickel + Dime}$$

Balancing equations as a step in research

Sometimes, we don't know the formula of a product yet, but we know what reactants were used. For example, if somebody serves me a concoction for desert. I don't know what it is, but it seems be made from eggs, apples and marshmallows (yuk!). I would write:

$$\ce{Egg + Apple + Marshmallow -> Concoction}$$

Then, as I do more research, I find out what recipe was used:

$$\ce{3 Egg + Apple + 5 Marshmallow -> Concoction}$$

What are possible misconceptions?

In a balanced equation, absence of a coefficient means that the coefficient is 1. In an unbalanced equation exercise, absence of a coefficient means that we have to find what it is. So the absence of coefficients does not necessarily means it needs to be balanced, e.g. $$\ce{NaCl(s) -> Na+(aq) + Cl-(aq)}$$

is balanced, while

$$\ce{MgCl2(s) -> Mg^2+(aq) + Cl-(aq)}$$

is not and does not represent anything that would actually happen. If students learning chemistry already had PhDs in math, we could pose the exercise simply as:

$$\ce{\nu_{MgCl_2} MgCl2(s) -> \nu_{Mg^{2+}} Mg^2+(aq) + \nu_{Cl^-} Cl-(aq)}$$

Determine the minimal integer values of nu (i.e. adding an additional constraint so the set of linear equations has a single solution in the simplest case). Omit the trivial solution of all values of nu vanishing. (Of course, the entire exercise is trivial if you have a PhDs in maths.)

In most cases, students learning chemistry don't have a PhD in math, so we avoid the greek symbols and keep the subscripts to a minimum.

$\endgroup$
  • $\begingroup$ Using the recipe analogy, I understand that when we're presented with an unbalanced equation exercise, we're actually being asked to find the minimum number of atoms needed in the reactants and the actual number of atoms in the product, such that this chemical reaction would be possible? That means that unbalanced equations are really just a generalized chemical reaction, not precise enough in terms of number of atoms. Am I getting this? $\endgroup$ – Pineapple29 Apr 27 at 13:52
  • 1
    $\begingroup$ Yes the stoichiomtric coefficients are the unknowns you are figuring out. $\endgroup$ – Karsten Theis Apr 27 at 14:43
5
$\begingroup$

First, provided the ball is an atom, you are not writing a chemical equation for a dozen of balls, rather a dozen times at least $10^{16}$ (to speak of equilibrium), usually a dozen times $10^{23}$ (macroscale that normal people operate on). That's not the level you want to use addition for; smaller numbers are easier to operate with.

Second, it's not always single "balls". Sometimes "balls" are "glued" together so that they cannot be separated prior to addition (the reactant is a polyatomic molecule). Sometimes it's a tremendously huge, but ordered set of "glued balls" (bulk solids, crystalline matter) so that counting the absolute value of all sorts of balls is nearly impossible, but the ratio between, lets say, green and yellow balls can be easily determined and, lets say, is exactly $1:1$ (e.g. sodium chloride NaCl).

Third, the "ball" can carry additional label that is also to be accounted when it comes to chemical reactions (e.g. charges and isotopes), whereas you can neglect those while counting the balls in Modell's Sporting Goods.

All these principles is the basis of stoichiometry.

$\endgroup$
1
$\begingroup$

Different compounds always have to have the same number of each type of atom in them. For example, in respiration:

$$\ce{C6H12O6 + O2 -> CO2 + H2O}$$

The chemical formulas for these compounds must always remain the same else it is no longer that compound. With this in mind balancing increases the number of molecules of each compound you have to ensure that there is the same number of atoms on each side of the equation, so it becomes

$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6H2O}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.