I don't believe you can use $\displaystyle \chi^2$ distribution for your test. As

this discussion details, among other things, the degree of freedom of the test corresponds to the number of cells in your analysis. Their example is to compare three distributions of fish, call them A, B, and C. They each have certain proportions a, b, and c. The expected proportions (E) were equally 1/3 for each cell. Thus, the test statistic is

$\displaystyle \chi^2 = \frac{(a - E)^2}{E} + \frac{(b - E)^2}{E} + \frac{(c - E)^2}{E}$

The degrees of freedom for this test is given by:

$\displaystyle df = "number\ of\ cells" - 1$

For the example, df = 2. In your case, you could only have a test of cell size = 1. Then your df = 0. How do you do a test with zero degrees of freedom? The answer is you cannot. Now, if you were to split your sample into males and females and have the population expected (empirical?) proportions for males and females. Then you could do the test with cell size = 2 for df = 1. The test is then straight-forward.